In the standard (x, y) coordinate plane, the line defined by the equation 3x - 4y = k is t… — ACT

In the standard $(x, y)$ coordinate plane, the line defined by the equation $3x - 4y = k$ is tangent to the circle defined by the equation $x^2 + y^2 - 2x - 4y = 4$ . If $k > 0$ , what is the value of $k$ ?

Cevap: 10

Cevap

The correct answer is 10. Completing the square for the circle's equation gives

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

, showing the center is

(1, 2)

and the radius is

3

. The distance from

(1, 2)

to the line

3x - 4y - k = 0

\frac{|3(1) - 4(2) - k|}{\sqrt{3^2 + (-4)^2}} = \frac{|k + 5|}{5}

. For tangency, this distance must equal the radius:

\frac{|k + 5|}{5} = 3

, which gives

|k + 5| = 15

. Solving this absolute value equation gives

k = 10

k = -20

. Since

k

must be positive, the value is 10.

Adım Adım Çözüm

Complete the square for the circle's equation

x^2 + y^2 - 2x - 4y = 4

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

, which represents a circle with center

(1, 2)

and radius

R = 3

To find the center and radius of the circle, which are needed to use the distance formula.

Express the distance

d

from the center

(1, 2)

to the line

3x - 4y - k = 0

using the formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

d = \frac{|3(1) - 4(2) - k|}{\sqrt{3^2 + (-4)^2}} = \frac{|-5 - k|}{5} = \frac{|k + 5|}{5}

A line is tangent to a circle if and only if the distance from the center of the circle to the line equals the radius.

Set the distance equal to the radius (

3

) and solve for

k

\frac{|k + 5|}{5} = 3 \Rightarrow |k + 5| = 15

, which yields

k + 5 = 15 \Rightarrow k = 10

, or

k + 5 = -15 \Rightarrow k = -20

To find the values of

k

that make the line tangent to the circle.

Select the positive value of

k

k = 10

The problem specifies that

k > 0

Anahtar Kavram

Systems of Linear and Non-Linear Equations

Tahmini Süre:2m 30s

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