In the -plane, the graph of the given equation is a circle. The line intersects the circle at two points. What is the distance between these two intersection points?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 101 | Free SAT Prep | Aniko

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Question 101·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the $xy$ -plane, the graph of the given equation is a circle. The line $y=2x-1$ intersects the circle at two points. What is the distance between these two intersection points? Аnіko‌.аi‍ -‍ ЅА‌T ‌Prep

x^2+y^2-6x+8y-11=0

For a circle-line intersection distance, rewrite the circle in standard form (so you’re working cleanly), substitute the line equation into the circle to get a quadratic in one variable, solve for the two intersection points, and then apply the distance formula. Keep expressions in exact radical form and simplify at the end. aniкo.аі

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Rewrite the circle

Complete the square to rewrite the circle as $(x-h)^2+(y-k)^2=r^2$ .

Use the intersection idea

Points of intersection satisfy both equations. Substitute $y=2x-1$ into the circle equation to get a quadratic in $x$ .

Distance between two points

After you find the two intersection points, use the distance formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ .

Desmos Guide

Graph the circle and the line

In Desmos, enter:

x^2 + y^2 - 6x + 8y - 11 = 0

and on a new line:

y = 2x - 1

Find the intersection points

Click on an intersection point and select the intersection coordinates. Do this for both intersection points. А-n-і‌-к-⁠o.аi

Compute the distance

Use Desmos to compute the distance with

d = sqrt((x1-x2)^2 + (y1-y2)^2)

where $(x1,y1)$ and $(x2,y2)$ are the two intersection points you found. Match the simplified value to the answer choices.

Step-by-step Explanation

Write the circle in standard form

Start by grouping terms and completing the square:

\begin{aligned} x^2-6x+y^2+8y&=11\\ (x-3)^2+(y+4)^2&=36 \end{aligned}

So the circle can be written as $(x-3)^2+(y+4)^2=36$ .

Substitute the line into the circle

Use $y=2x-1$ in $(x-3)^2+(y+4)^2=36$ :

(x-3)^2+\big((2x-1)+4\big)^2=36

which simplifies to

(x-3)^2+(2x+3)^2=36.

Solve for the intersection $x$ -coordinates

Expand and simplify:

\begin{aligned} (x^2-6x+9)+(4x^2+12x+9)&=36\\ 5x^2+6x+18&=36\\ 5x^2+6x-18&=0 \end{aligned}

Solve the quadratic: (Anік⁠o.аi‌)

\begin{aligned} x&=\frac{-6\pm\sqrt{6^2-4(5)(-18)}}{2\cdot 5} =\frac{-6\pm\sqrt{396}}{10} =\frac{-6\pm 6\sqrt{11}}{10} =\frac{-3\pm 3\sqrt{11}}{5}. \end{aligned}

Find the two intersection points and set up the distance

For each $x$ , compute $y=2x-1$ .

If $x_1=\frac{-3+3\sqrt{11}}{5}$ , then

y_1=2\left(\frac{-3+3\sqrt{11}}{5}\right)-1=\frac{-11+6\sqrt{11}}{5}.

If $x_2=\frac{-3-3\sqrt{11}}{5}$ , then

y_2=2\left(\frac{-3-3\sqrt{11}}{5}\right)-1=\frac{-11-6\sqrt{11}}{5}.

Now compute the differences:

\Delta x=x_1-x_2=\frac{6\sqrt{11}}{5},\qquad \Delta y=y_1-y_2=\frac{12\sqrt{11}}{5}.

Use the distance formula and simplify

The distance between the intersection points is

\begin{aligned} d&=\sqrt{(\Delta x)^2+(\Delta y)^2} =\sqrt{\left(\frac{6\sqrt{11}}{5}\right)^2+\left(\frac{12\sqrt{11}}{5}\right)^2}\\ &=\sqrt{\frac{36\cdot 11}{25}+\frac{144\cdot 11}{25}} =\sqrt{\frac{180\cdot 11}{25}} =\frac{\sqrt{1980}}{5} =\frac{6\sqrt{55}}{5}. \end{aligned}

Therefore, the correct choice is $\frac{6\sqrt{55}}{5}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation