A circle and a line are graphed in the standard (x, y) coordinate plane. The equations of … — ACT

A circle and a line are graphed in the standard $(x, y)$ coordinate plane. The equations of the circle and the line are given by:

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

y = x + 1

The line intersects the circle at two points, $P$ and $Q$ . What is the distance between $P$ and $Q$ ?

The distance between the intersection points is

3\sqrt{2}

To find the points of intersection, substitute the linear equation into the circle's equation. This results in the quadratic equation

x^2 - 5x + 4 = 0

, which yields the solutions

x = 1

and

x = 4

. Substituting these values into the linear equation gives the points

(1, 2)

and

(4, 5)

. The distance between these points is computed using the distance formula, which gives

3\sqrt{2}

Substitute the equation of the line into the equation of the circle.

(x-3)^2 + ((x+1)-3)^2 = 5 \implies (x-3)^2 + (x-2)^2 = 5

This reduces the system of two equations with two variables to a single quadratic equation in terms of

x

Expand the squared binomials and simplify the quadratic equation.

(x^2 - 6x + 9) + (x^2 - 4x + 4) = 5 \implies 2x^2 - 10x + 13 = 5 \implies 2x^2 - 10x + 8 = 0 \implies x^2 - 5x + 4 = 0

Expanding the terms allows us to combine like terms and set the quadratic equation to zero.

Factor the quadratic equation to find the

x

-coordinates of the intersection points.

(x-1)(x-4) = 0 \implies x = 1 \text{ or } x = 4

Factoring is the most direct method to solve the simplified quadratic equation.

Find the corresponding

y

-coordinates by substituting the

x

-values back into the linear equation

y = x + 1

\text{For } x = 1: y = 1 + 1 = 2 \implies P(1, 2)

\text{For } x = 4: y = 4 + 1 = 5 \implies Q(4, 5)

This determines the coordinates of the two intersection points.

Use the distance formula to find the distance between the two points

P(1, 2)

and

Q(4, 5)

d = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}

The distance formula calculates the straight-line distance between the two coordinates.

Solving systems of linear and quadratic equations by substitution and finding the distance between intersection points.

Find the distance geometrically: The center of the circle is

(3, 3)

and the radius is

r = \sqrt{5}

. The distance

d

from the center to the line

x - y + 1 = 0

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

. Using a right triangle formed by the radius, the distance from the center, and half of the chord length

h

, we have

h = \sqrt{r^2 - d^2} = \sqrt{5 - \frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}

. The total distance between the intersection points is the full chord length,

2h = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}

Estimated Time:3m 0s

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