In the standard (x, y) coordinate plane, the circle defined by x^2 + y^2 = 26 and the line… — ACT

In the standard $(x, y)$ coordinate plane, the circle defined by $x^2 + y^2 = 26$ and the line defined by $y = x - 4$ intersect at two points. What is the distance between these two points of intersection?

A
$2\sqrt{10}$
B
$4\sqrt{2}$
C
$6$
$6\sqrt{2}$ Answer
E
$12$

Answer

6\sqrt{2}

Solving the system of equations by substituting

y = x - 4

into

x^2 + y^2 = 26

yields

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

. Factoring gives

(x-5)(x+1) = 0

, leading to the points

(5, 1)

and

(-1, -5)

. The distance between these points is

\sqrt{6^2 + 6^2} = 6\sqrt{2}

Step-by-Step Solution

Substitute the linear equation into the circle equation.

x^2 + (x - 4)^2 = 26

This reduces the system to a single equation in terms of

x

Expand the squared binomial and simplify the quadratic equation.

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

Expanding

(x - 4)^2

yields

x^2 - 8x + 16

. Setting the equation to zero allows us to solve for

x

Factor the quadratic equation to find the

x

-coordinates.

x = 5

x = -1

The factored form is

(x - 5)(x + 1) = 0

Substitute the

x

-values back into the linear equation to find the corresponding

y

-coordinates.

The intersection points are

(5, 1)

and

(-1, -5)

For

x = 5

y = 5 - 4 = 1

. For

x = -1

y = -1 - 4 = -5

Apply the distance formula to find the straight-line distance between the two points.

6\sqrt{2}

The distance is

\sqrt{(5 - (-1))^2 + (1 - (-5))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}

Key Concept

Solving systems of linear and non-linear (circular) equations by substitution and finding the distance between intersection points.

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