SAT Circles Question 32 | Free SAT Prep | Aniko

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Question 32·Hard·Circles

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A circle in the $xy$ -plane has equation

x^2 + y^2 - 4x + 6y - 12 = 0.

From the point $(7,1)$ , two distinct lines can be drawn that are tangent to the circle. One of these tangents is the vertical line $x = 7$ . Which of the following is the slope of the other tangent line?

For tangent-line-from-a-point-to-a-circle questions, first rewrite the circle in $(x-h)^2 + (y-k)^2 = r^2$ form to quickly read off the center and radius. Then write the general equation of a line through the given point with unknown slope $m$ , convert it to standard form, and use the point-to-line distance formula so that the distance from the center to this line equals the radius. This gives one equation in one variable $m$ , which you can solve algebraically; this approach is usually faster and less error-prone on the SAT than trying to find intersection points and using the quadratic discriminant.

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Identify the circle's center and radius

Rewrite $x^2 + y^2 - 4x + 6y - 12 = 0$ in the form $(x-h)^2 + (y-k)^2 = r^2$ by completing the square in $x$ and in $y$ . What are $h$ , $k$ , and $r$ ?

Describe all possible lines through (7,1)

Any non-vertical line through $(7,1)$ can be written using slope $m$ as $y - 1 = m(x - 7)$ . Use this as the general form of the other tangent line.

Use a distance condition for tangency

For the line you just wrote, convert it to $Ax + By + C = 0$ and use the formula for the distance from a point to a line. Set that distance (from the circle's center) equal to the circle's radius.

Solve the resulting equation

After you set up the distance equation, you will have an equation involving $m$ with a square root and absolute value. Square both sides carefully and solve for $m$ , remembering that the vertical tangent $x = 7$ is a separate case and already given.

Desmos Guide

Graph the circle

Enter the circle in center-radius form: (x-2)^2 + (y+3)^2 = 25. This will show a circle centered at $(2,-3)$ with radius $5$ .

Graph a family of lines through (7,1)

Create a slider m in Desmos, then type y - 1 = m(x - 7). As you move the m slider, this line will pivot around the point $(7,1)$ , showing all possible non-vertical lines through that point.

Find which line is tangent

Adjust the m slider until the line just touches the circle at exactly one point (it should intersect the circle in a single point instead of crossing through). Read the value of m from the slider; that value is the slope of the non-vertical tangent. You can then match that numeric value to the closest answer choice.

Step-by-step Explanation

Find the center and radius of the circle

Start with the given equation:

x^2 + y^2 - 4x + 6y - 12 = 0.

Complete the square in $x$ and $y$ :

\begin{aligned} x^2 - 4x + y^2 + 6y - 12 &= 0 \\ (x-2)^2 - 4 + (y+3)^2 - 9 - 12 &= 0 \\ (x-2)^2 + (y+3)^2 &= 25. \end{aligned}

So the circle has center $(2,-3)$ and radius $r = 5$ .

Write the equation of a general line through (7,1)

Let the slope of the other tangent line (not the vertical one) be $m$ .

Any non-vertical line through $(7,1)$ can be written as

y - 1 = m(x - 7).

Rewrite this in standard form $Ax + By + C = 0$ so we can use the distance formula:

\begin{aligned} y - 1 &= m(x - 7) \\ y - 1 &= mx - 7m \\ mx - y - 7m + 1 &= 0. \end{aligned}

So for this line, $A = m$ , $B = -1$ , and $C = -7m + 1$ .

Use the distance formula from the center to the line

For a line $Ax + By + C = 0$ and a point $(x_0,y_0)$ , the distance from the point to the line is

\text{distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.

Here, the center is $(2,-3)$ and the line is $mx - y - 7m + 1 = 0$ .

Compute the numerator:

\begin{aligned} Ax_0 + By_0 + C &= m(2) + (-1)(-3) + (-7m + 1) \\ &= 2m + 3 - 7m + 1 \\ &= -5m + 4. \end{aligned}

So the distance from the center to the line is

\frac{| -5m + 4 |}{\sqrt{m^2 + 1}}.

Because the line is tangent, this distance must equal the radius $5$ :

\frac{| -5m + 4 |}{\sqrt{m^2 + 1}} = 5.

Solve for the slope m

Square both sides to remove the absolute value and the square root:

\frac{(-5m + 4)^2}{m^2 + 1} = 25.

Cross-multiply:

(-5m + 4)^2 = 25(m^2 + 1).

Expand both sides:

\begin{aligned} (-5m + 4)^2 &= 25m^2 - 40m + 16, \\ 25(m^2 + 1) &= 25m^2 + 25. \end{aligned}

Set them equal and simplify:

\begin{aligned} 25m^2 - 40m + 16 &= 25m^2 + 25 \\ -40m + 16 &= 25 \\[-2pt] -40m &= 9 \\ m &= -\frac{9}{40}. \end{aligned}

This is the slope of the non-vertical tangent line through $(7,1)$ , so the correct choice is $-\dfrac{9}{40}$ .

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