Coordinate Geometry Alıştırma Soruları — ACT — Sayfa 14

Soru 261Soru →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 12x + 6y + 20 = 0$ . What is the distance from the center of this circle to the point $(9, 1)$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The distance from the center of the circle to the point

(9, 1)

is 5.

Completing the square transforms

x^2 + y^2 - 12x + 6y + 20 = 0

into standard form

(x - 6)^2 + (y + 3)^2 = 25

, establishing the center of the circle at

(6, -3)

. Applying the coordinate distance formula between

(6, -3)

and

(9, 1)

yields

\sqrt{(9 - 6)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

.

Adım Adım Çözüm

1

Group the

x

and

y

terms and move the constant term to the right side of the equation.

(x^2 - 12x) + (y^2 + 6y) = -20

Grouping terms isolates the variables to prepare for completing the square.

2

Complete the square for both variable expressions.

(x - 6)^2 + (y + 3)^2 = -20 + 36 + 9 = 25

Adding

( -12 / 2 )^2 = 36

and

( 6 / 2 )^2 = 9

to both sides converts the equation to standard form

(x - h)^2 + (y - k)^2 = r^2

.

3

Determine the coordinates of the center

(h, k)

.

Center is

(6, -3)

In the standard circle equation

(x - h)^2 + (y - k)^2 = r^2

, the center is given by

(h, k)

.

4

Calculate the distance between the center

(6, -3)

and the point

(9, 1)

using the distance formula.

d = \sqrt{(9 - 6)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

The distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

measures the straight-line distance between two points.

Anahtar Kavram

Equations of Circles and Coordinate Distance

Soru 262Soru →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 8x + 10y + k = 0$ , where $k$ is a constant. If the circle is tangent to the $y$ -axis, what is the value of $k$ ?

16

25

36

37

41

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The value of

k

is 25.

Completing the square for

x^2 + y^2 - 8x + 10y + k = 0

gives

(x - 4)^2 + (y + 5)^2 = 41 - k

. The center of the circle is

(4, -5)

. Because the circle is tangent to the

y

-axis, its radius is equal to the absolute value of the center's

x

-coordinate, which is

|4| = 4

. Therefore,

r^2 = 16

. Setting

41 - k = 16

gives

k = 25

.

Adım Adım Çözüm

1

Rewrite the general equation of the circle by completing the square for both

x

and

y

.

(x^2 - 8x + 16) + (y^2 + 10y + 25) + k - 16 - 25 = 0 \implies (x - 4)^2 + (y + 5)^2 = 41 - k

Converting to standard form

(x - h)^2 + (y - k)^2 = r^2

reveals the center

(h, k)

and radius squared

r^2

.

2

Identify the center of the circle and determine the radius using the tangency condition.

Center is

(4, -5)

. Since the circle is tangent to the

y

-axis (the line

x = 0

), the radius

r

equals the horizontal distance from the center to the

y

-axis:

r = |4| = 4

.

Tangency to a vertical line means the distance from the center to that line is equal to the radius.

3

Equate the radius squared from step 1 to

r^2

from step 2 and solve for

k

.

41 - k = 4^2 \implies 41 - k = 16 \implies k = 25

The right side of the standard circle equation represents

r^2

.

Anahtar Kavram

Standard form of a circle equation and conditions for tangency to coordinate axes

Soru 263Soru →

A circle in the standard $(x, y)$ coordinate plane is represented by the equation $x^2 + y^2 + 4x - 6y - 12 = 0$ . What is the distance between the two points where this circle intersects the $x$ -axis?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The distance between the two points where the circle intersects the

x

-axis is 8 units.

Setting

y = 0

gives the quadratic equation

x^2 + 4x - 12 = 0

. Factoring yields

(x + 6)(x - 2) = 0

, giving solutions

x = -6

and

x = 2

. The distance between these two

x

-intercepts is

2 - (-6) = 8

units.

Adım Adım Çözüm

1

Substitute

y = 0

into the circle equation.

x^2 + 4x - 12 = 0

Points on the

x

-axis have a

y

-coordinate of 0.

2

Solve the quadratic equation for

x

by factoring.

x = -6

and

x = 2

Factoring

(x + 6)(x - 2) = 0

yields the

x

-coordinates of the intersection points.

3

Find the distance between

(-6, 0)

and

(2, 0)

.

2 - (-6) = 8

The horizontal distance between two points

(x_1, 0)

and

(x_2, 0)

is

|x_2 - x_1|

.

Anahtar Kavram

Finding

x

-intercepts of a circle given in general form

Soru 264Soru →

A circle in the standard $(x, y)$ coordinate plane is described by the equation $x^2 + y^2 - 6x + 10y + 18 = 0$ . A second circle is concentric with the first circle (meaning they share the exact same center) but has an area that is $4$ times the area of the first circle. Which of the following is an equation of the second circle?

(x - 3)^2 + (y + 5)^2 = 64

(x + 3)^2 + (y - 5)^2 = 64

(x - 3)^2 + (y + 5)^2 = 32

(x + 3)^2 + (y - 5)^2 = 32

(x - 3)^2 + (y + 5)^2 = 16

Cevabı ve açıklamayı göster

Cevap: $(x - 3)^2 + (y + 5)^2 = 64$

Cevap

The equation of the second circle is

(x - 3)^2 + (y + 5)^2 = 64

.

Completing the square on

x^2 + y^2 - 6x + 10y + 18 = 0

yields

(x - 3)^2 + (y + 5)^2 = 16

, identifying the center as

(3, -5)

and

r_1^2 = 16

. Concentric circles share the center

(3, -5)

. Quadrupling the area means the new area is

4 \times 16\pi = 64\pi

, so

r_2^2 = 64

. Plugging the center and new

r^2

into the standard equation gives

(x - 3)^2 + (y + 5)^2 = 64

.

Adım Adım Çözüm

1

Convert the given circle equation to standard form by completing the square for both

x

and

y

.

Rearranging terms:

(x^2 - 6x) + (y^2 + 10y) = -18

.
Adding

( -6/2 )^2 = 9

and

( 10/2 )^2 = 25

to both sides yields:

(x - 3)^2 + (y + 5)^2 = -18 + 9 + 25 = 16

.

Standard form

(x - h)^2 + (y - k)^2 = r^2

reveals the center

(h, k)

and squared radius

r^2

.

2

Identify the center and radius of the original circle.

Center =

(3, -5)

and radius squared

r_1^2 = 16

(so

r_1 = 4

).

Concentric circles share the exact same center

(h, k) = (3, -5)

.

3

Determine the radius squared of the second circle based on the area constraint.

The area of a circle is

A = \pi r^2

. Since

A_2 = 4 A_1

, we have

\pi r_2^2 = 4 (\pi r_1^2) \implies r_2^2 = 4 r_1^2 = 4(16) = 64

.

Quadrupling the area quadruples the value of

r^2

.

4

Write the standard form equation for the second circle using center

(3, -5)

and

r_2^2 = 64

.

(x - 3)^2 + (y + 5)^2 = 64

.

Substituting

h = 3

,

k = -5

, and

r_2^2 = 64

into

(x - h)^2 + (y - k)^2 = r_2^2

.

Anahtar Kavram

Standard form of a circle equation

(x - h)^2 + (y - k)^2 = r^2

and area relationship

A = \pi r^2

Soru 265Soru →

In the standard $(x, y)$ coordinate plane, a circle is tangent to the horizontal line $y = 7$ . The equation of the circle is $x^2 + y^2 - 10x + 4y + c = 0$ , where $c$ is a constant. What is the value of $c$ ?

Cevabı ve açıklamayı göster

Cevap: -52

Cevap

The value of

c

is

-52

.

Completing the square gives

(x - 5)^2 + (y + 2)^2 = 29 - c

. The center is

(5, -2)

. The distance from

(5, -2)

to the horizontal line

y = 7

is

|7 - (-2)| = 9

, which is the radius

r

. Squaring the radius gives

r^2 = 81

. Setting

29 - c = 81

gives

c = -52

.

Adım Adım Çözüm

1

Complete the square for

x

and

y

terms.

(x - 5)^2 + (y + 2)^2 = 29 - c

Writing the equation in standard form

(x - h)^2 + (y - k)^2 = r^2

reveals the center

(h, k)

and the expression for

r^2

.

2

Calculate the radius using the distance from the center to the tangent line.

Radius

r = 9

The center is

(5, -2)

and the tangent line is

y = 7

. The distance between

y = -2

and

y = 7

is

|7 - (-2)| = 9

.

3

Set

r^2 = 81

and solve for

c

.

c = -52

Equating the radius squared expression

29 - c

to

81

yields

29 - c = 81 \implies c = -52

.

Anahtar Kavram

Converting circle equations to standard form and using distance to tangent lines to determine radius.

Soru 266Soru →

A circle in the standard $(x, y)$ coordinate plane is represented by the equation $x^2 + y^2 + 10x - 6y + 9 = 0$ . If this circle is translated $4$ units to the right and $2$ units down, which of the following is the equation of the translated circle?

(x + 1)^2 + (y - 1)^2 = 25

(x - 1)^2 + (y + 1)^2 = 25

(x + 1)^2 + (y - 1)^2 = 5

(x + 9)^2 + (y - 5)^2 = 25

(x + 1)^2 + (y - 1)^2 = 9

Cevabı ve açıklamayı göster

Cevap: $(x + 1)^2 + (y - 1)^2 = 25$

Cevap

(x + 1)^2 + (y - 1)^2 = 25

Completing the square on

x^2 + y^2 + 10x - 6y + 9 = 0

yields

(x + 5)^2 + (y - 3)^2 = 25

, identifying the original center as

(-5, 3)

and

r^2 = 25

. Shifting the center

4

units right and

2

units down moves

(-5, 3)

to

(-5 + 4, 3 - 2) = (-1, 1)

. Substituting the updated center

(-1, 1)

and

r^2 = 25

into the standard equation of a circle

(x - h)^2 + (y - k)^2 = r^2

gives

(x + 1)^2 + (y - 1)^2 = 25

.

Adım Adım Çözüm

1

Group terms and complete the square for both

x

and

y

variables in the original equation.

(x + 5)^2 + (y - 3)^2 = 25

Rearrange

x^2 + y^2 + 10x - 6y + 9 = 0

as

(x^2 + 10x) + (y^2 - 6y) = -9

. Add

(10/2)^2 = 25

and

(-6/2)^2 = 9

to both sides:

(x + 5)^2 + (y - 3)^2 = -9 + 25 + 9 = 25

.

2

Determine the center and radius of the original circle.

Original center is

(-5, 3)

and

r^2 = 25

(radius

r = 5

).

Comparing

(x + 5)^2 + (y - 3)^2 = 25

to

(x - h)^2 + (y - k)^2 = r^2

gives

h = -5

,

k = 3

, and

r^2 = 25

.

3

Calculate the coordinates of the new center after the horizontal and vertical translation.

New center is

(-1, 1)

.

Translating

4

units to the right adds

4

to the

x

-coordinate:

-5 + 4 = -1

. Translating

2

units down subtracts

2

from the

y

-coordinate:

3 - 2 = 1

.

4

Substitute the new center

(-1, 1)

and unchanged radius squared value

r^2 = 25

into standard form.

(x + 1)^2 + (y - 1)^2 = 25

Using

(x - h)^2 + (y - k)^2 = r^2

with

h = -1

and

k = 1

gives

(x - (-1))^2 + (y - 1)^2 = 25

, which simplifies to

(x + 1)^2 + (y - 1)^2 = 25

.

Anahtar Kavram

Converting circle equations to standard form by completing the square and performing translations in the coordinate plane.

Soru 267Soru →

In the standard $(x, y)$ coordinate plane, the equation of a circle is given by $3x^2 + 3y^2 - 12x + 24y - 15 = 0$ . What is the radius of this circle?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The radius of the circle is 5 units.

Dividing the equation

3x^2 + 3y^2 - 12x + 24y - 15 = 0

by 3 gives

x^2 + y^2 - 4x + 8y - 5 = 0

. Completing the square for both variables results in

(x - 2)^2 + (y + 4)^2 = 25

. Because

r^2 = 25

, taking the square root gives the radius

r = 5

.

Adım Adım Çözüm

1

Divide the entire equation by 3.

x^2 + y^2 - 4x + 8y - 5 = 0

The standard form of a circle equation

(x-h)^2 + (y-k)^2 = r^2

requires the coefficients of

x^2

and

y^2

to be 1.

2

Group terms and move the constant to the right side of the equation.

(x^2 - 4x) + (y^2 + 8y) = 5

Isolating quadratic terms makes completing the square straightforward.

3

Complete the square for both variables by adding

(\frac{-4}{2})^2 = 4

and

(\frac{8}{2})^2 = 16

to both sides.

(x^2 - 4x + 4) + (y^2 + 8y + 16) = 5 + 4 + 16

Adding these values turns the expressions into perfect square trinomials while preserving equality.

4

Express in standard form and solve for the radius

r

.

(x - 2)^2 + (y + 4)^2 = 25 \implies r^2 = 25 \implies r = 5

The radius is the principal square root of

r^2

.

Anahtar Kavram

Converting general form of a circle equation to standard form using completing the square.

Soru 268Soru →

An ellipse in the standard $(x, y)$ coordinate plane is defined by the equation

\frac{(x-2)^2}{9} + \frac{(y+5)^2}{16} = 1

What are the coordinates of the center of this ellipse?

(-2, 5)

(2, 5)

(2, -5)

(-2, -5)

(-5, 2)

Cevabı ve açıklamayı göster

Cevap: $(2, -5)$

Cevap

The center of the ellipse is

(2, -5)

.

The standard form of an ellipse equation is

\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

, where the center of the ellipse is at the coordinate point

(h, k)

. Comparing the given equation

\frac{(x-2)^2}{9} + \frac{(y+5)^2}{16} = 1

to the standard form reveals that

h = 2

and

k = -5

. Therefore, the center coordinates are

(2, -5)

.

Adım Adım Çözüm

1

Recall the standard form equation of an ellipse with horizontal/vertical axes.

The standard form is

\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

, where

(h, k)

represents the coordinates of the center.

This establishes the framework to extract the center coordinate values.

2

Compare the terms in the given equation to the standard form.

Matching

(x-h)^2

with

(x-2)^2

gives

h = 2

. Matching

(y-k)^2

with

(y+5)^2 = (y-(-5))^2

gives

k = -5

.

Comparing terms identifies the offsets

h

and

k

that determine the center.

3

Write the center coordinate pair

(h, k)

.

The center is

(2, -5)

.

Combining the values of

h

and

k

yields the final coordinates.

Anahtar Kavram

Identifying the center of an ellipse from its standard form equation

Tahmini Süre:45s

Soru 269Soru →

In the standard $(x, y)$ coordinate plane, line $l_1$ passes through the points $(-2, 5)$ and $(4, 1)$ . Line $l_2$ is perpendicular to line $l_1$ . What is the slope of line $l_2$ ?

-\frac{3}{2}

-\frac{2}{3}

\frac{2}{3}

\frac{3}{2}

\frac{1}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{3}{2}$

Cevap

The slope of the perpendicular line is

\frac{3}{2}

.

To find the slope of a line perpendicular to a given line, first calculate the slope of the original line,

l_1

, using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting the points

(-2, 5)

and

(4, 1)

into the formula gives

m_1 = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}

. The slope of a perpendicular line,

l_2

, is the negative reciprocal of the slope of

l_1

. The negative reciprocal of

-\frac{2}{3}

is

\frac{3}{2}

.

Adım Adım Çözüm

1

Calculate the slope of the line

l_1

passing through the points

(-2, 5)

and

(4, 1)

using the slope formula.

m_1 = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}

The slope

m

of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Find the slope of line

l_2

, which is perpendicular to

l_1

, by taking the negative reciprocal of

m_1

.

m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}

The product of the slopes of two perpendicular lines is

-1

, so the slope of a perpendicular line is the negative reciprocal of the original slope.

Anahtar Kavram

Slope of perpendicular lines and finding slope from two points.

Tahmini Süre:1m 0s

Soru 270Soru →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $A(0, 1)$ , $B(2, 3)$ , and $C(8, 11)$ . If point $M$ is the midpoint of side $AB$ and point $N$ is the midpoint of side $AC$ , what is the length of the line segment $MN$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The length of the line segment

MN

is

5

.

The length of the line segment

MN

is

5

. Calculating the coordinates of the midpoint of

AB

, we get

M(1, 2)

. For

AC

, the midpoint is

N(4, 6)

. Applying the distance formula between

M

and

N

yields

\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9+16} = 5

. Alternatively, by the Midsegment Theorem, the segment connecting the midpoints of two sides of a triangle is half the length of the third side. The length of the third side

BC

is

\sqrt{(8-2)^2 + (11-3)^2} = \sqrt{36 + 64} = 10

, so the length of

MN

is

\frac{10}{2} = 5

.

Adım Adım Çözüm

1

Find the coordinates of

M

, the midpoint of side

AB

with endpoints

A(0, 1)

and

B(2, 3)

.

M(1, 2)

The midpoint formula is

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

.

2

Find the coordinates of

N

, the midpoint of side

AC

with endpoints

A(0, 1)

and

C(8, 11)

.

N(4, 6)

Applying the midpoint formula gives

\left(\frac{0 + 8}{2}, \frac{1 + 11}{2}\right) = (4, 6)

.

3

Calculate the distance between

M(1, 2)

and

N(4, 6)

using the distance formula.

5

The distance formula is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

. Here,

d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

.

Anahtar Kavram

Midpoint and Distance Formulas

Soru 271Soru →

A vertex of a polygon in the standard $(x, y)$ coordinate plane undergoes two transformations: it is first reflected across the line $y = x$ , and then it is translated $4$ units to the right. If the coordinates of the final image of the vertex are $(1, 5)$ , what were the coordinates of the vertex before the transformations?

(1, 1)

(5, 5)

(5, -3)

(9, 1)

(-3, 5)

Cevabı ve açıklamayı göster

Cevap: $(5, -3)$

Cevap

The coordinates of the vertex before the transformations were

(5, -3)

.

To find the pre-image of the vertex, we apply the inverse transformations in reverse order. First, we undo the translation of

4

units to the right by translating the image point

(1, 5)

by

4

units to the left, resulting in

(-3, 5)

. Second, we undo the reflection across the line

y = x

by swapping the coordinates of

(-3, 5)

, which yields the original point

(5, -3)

.

Adım Adım Çözüm

1

Identify the inverse of the second transformation (translation of

4

units to the right).

The inverse is a translation of

4

units to the left, which subtracts

4

from the

x

-coordinate:

(1 - 4, 5) = (-3, 5)

.

To find the pre-image, we must undo the transformations in reverse order, starting with the last transformation applied.

2

Identify the inverse of the first transformation (reflection across the line

y = x

).

The inverse of a reflection across

y = x

is itself, which swaps the

x

- and

y

-coordinates:

(-3, 5) \rightarrow (5, -3)

.

Undoing the first transformation on the intermediate coordinates gives the original pre-image coordinates.

Anahtar Kavram

Working backward from a final image using inverse transformations in reverse order.

Alternatif Yöntem

We can write the composite transformation as an algebraic rule. A reflection across

y = x

maps

(x, y) \to (y, x)

. A translation

4

units to the right maps

(y, x) \to (y + 4, x)

. Setting the final coordinates equal to

(1, 5)

, we get

y + 4 = 1

(which means

y = -3

) and

x = 5

. Thus, the original coordinates were

(5, -3)

.

Tahmini Süre:1m 15s

Soru 272Soru →

The equation $x^2 + y^2 - 12x + 8y + 3 = 0$ defines a circle in the standard $(x, y)$ coordinate plane. What is the radius of this circle?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The radius of the circle is 7.

Completing the square on the given equation yields

(x - 6)^2 + (y + 4)^2 = 49

. By comparing this to the standard circle equation

(x - h)^2 + (y - k)^2 = r^2

, we find

r^2 = 49

. Taking the square root of

49

gives a radius of

7

.

Adım Adım Çözüm

1

Group the

x

and

y

terms and move the constant to the right-hand side of the equation.

x^2 - 12x + y^2 + 8y = -3

This groups terms containing the same variables together to prepare for completing the square.

2

Add the square of half of the linear coefficients to both sides to complete the square for both

x

and

y

.

(x^2 - 12x + 36) + (y^2 + 8y + 16) = -3 + 36 + 16

Adding

(\frac{-12}{2})^2 = 36

and

(\frac{8}{2})^2 = 16

to both sides forms perfect square trinomials on the left side while maintaining equality.

3

Factor the trinomials into squared binomials and combine the constants on the right side.

(x - 6)^2 + (y + 4)^2 = 49

This rewrites the equation in the standard circle equation form.

4

Extract the radius from the standard form equation

(x - h)^2 + (y - k)^2 = r^2

.

r = 7

Since the constant on the right side corresponds to

r^2

, taking the square root of

49

gives the radius of the circle.

Anahtar Kavram

Converting a circle's equation from general form to standard form by completing the square to determine its properties.

Soru 273Soru →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 + 14x - 6y + 9 = 0$ . What is the area, in square units, of the region enclosed by the circle?

7\pi

14\pi

40\pi

49\pi

58\pi

Cevabı ve açıklamayı göster

Cevap: $49\pi$

Cevap

The area of the region enclosed by the circle is

49\pi

square units.

To find the area of the circle, we rewrite the equation

x^2 + y^2 + 14x - 6y + 9 = 0

in standard form

(x - h)^2 + (y - k)^2 = r^2

. Regrouping gives

(x^2 + 14x) + (y^2 - 6y) = -9

. Completing the square for

x

requires adding

(\frac{14}{2})^2 = 49

, and for

y

requires adding

(\frac{-6}{2})^2 = 9

. Adding both to the right side gives

-9 + 49 + 9 = 49

. Thus, the standard equation is

(x + 7)^2 + (y - 3)^2 = 49

, yielding

r^2 = 49

. The area enclosed by the circle is

\pi r^2 = 49\pi

.

Adım Adım Çözüm

1

Group terms and move the constant term to the right side of the equation.

(x^2 + 14x) + (y^2 - 6y) = -9

Grouping

x

-terms and

y

-terms prepares the equation for completing the square.

2

Complete the square for both

x

and

y

expressions by adding

(\frac{14}{2})^2 = 49

and

(\frac{-6}{2})^2 = 9

to both sides.

(x^2 + 14x + 49) + (y^2 - 6y + 9) = -9 + 49 + 9

Adding these values turns the quadratic expressions into perfect square trinomials while keeping the equation balanced.

3

Write in standard circle equation form

(x - h)^2 + (y - k)^2 = r^2

to identify

r^2

.

(x + 7)^2 + (y - 3)^2 = 49

, so

r^2 = 49

The standard form directly gives the radius squared,

r^2 = 49

.

4

Calculate the area of the circle using the formula

A = \pi r^2

.

A = \pi (49) = 49\pi

Substituting

r^2 = 49

into the area formula gives the final enclosed area.

Anahtar Kavram

Converting general circle equation

x^2 + y^2 + Dx + Ey + F = 0

to standard form

(x - h)^2 + (y - k)^2 = r^2

by completing the square.

Coordinate Geometry

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