SAT Circles Question 9 | Free SAT Prep | Aniko

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Question 9·Hard·Circles

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(x - 3)^2 + (y + 2)^2 = 50

In the $xy$ -plane, the graph of the equation above is a circle. Which of the following is an equation of a line with slope $4$ that is tangent to the circle?

For circle–line tangency questions, first rewrite the circle in center-radius form to read off the center and radius quickly. Then write a general line with the given slope, convert it to standard form, and use the perpendicular distance formula from the center to this line; set that distance equal to the radius and solve for the unknown intercept. Finally, compare your resulting line equation(s) with the answer choices, focusing on both the slope and the constant term so you can eliminate distractors that either cut through the circle or miss it entirely.

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Start from the circle’s standard form

Rewrite the given equation in the form $(x-h)^2 + (y-k)^2 = r^2$ . What are the center $(h,k)$ and the radius $r$ ?

Write a general line with the given slope

Any line with slope $4$ can be written as $y = 4x + b$ . Think of $b$ as the unknown you need to determine so the line is tangent to the circle.

Relate tangency to distance

A tangent line touches the circle at exactly one point. How is the distance from the center of the circle to the line related to the radius in that case?

Use the point-to-line distance formula

Use the formula for distance from a point to a line: for point $(x_0,y_0)$ and line $Ax + By + C = 0$ , the distance is $\dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ . Plug in the circle’s center and set this distance equal to the radius, then solve for $b$ .

Desmos Guide

Graph the circle

Type (x-3)^2 + (y+2)^2 = 50 into Desmos to graph the circle and visually confirm its center and size.

Graph each answer choice line

Enter each option as a separate line: y = 4x - 14, y = 4x - 14 - 2sqrt(34), y = 4x - 9 + 5sqrt(34), and y = 4x - 14 + 5sqrt(34).

Check how each line intersects the circle

For each line, look at how it interacts with the circle: one line will pass through the center, another will cross the circle at two points, another will miss the circle completely, and one line will just touch the circle at exactly one point (a single intersection).

Identify the tangent line

Use Desmos’s intersection tool (or click on the graphs) to see the number of intersection points for each line with the circle; the line that has exactly one intersection point with the circle is the tangent line, which corresponds to the correct answer choice.

Step-by-step Explanation

Identify the circle’s center and radius

The equation of a circle in standard form is

(x-h)^2 + (y-k)^2 = r^2,

where $(h,k)$ is the center and $r$ is the radius.

Compare this with

(x - 3)^2 + (y + 2)^2 = 50.

So:

Center: $(h,k) = (3,-2)$ (because $y+2$ means $y - (-2)$ ),
Radius: $r = \sqrt{50} = 5\sqrt{2}$ , since $50 = 25\cdot 2$ .

Write the general line with slope 4

A line with slope $4$ can be written as

y = 4x + b,

where $b$ is the $y$ -intercept.

To use the distance formula for a point to a line, rewrite this in standard form $Ax + By + C = 0$ :

4x - y + b = 0.

Here, $A = 4$ , $B = -1$ , and $C = b$ .

Use the perpendicular distance formula

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

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

Our point is the center $(3,-2)$ and the line is $4x - y + b = 0$ .

Compute the numerator:

A x_0 + B y_0 + C = 4(3) + (-1)(-2) + b = 12 + 2 + b = 14 + b.

The denominator is

\sqrt{A^2 + B^2} = \sqrt{4^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}.

So the distance from the center to the line $y = 4x + b$ is

\frac{|14 + b|}{\sqrt{17}}.

For the line to be tangent to the circle, this distance must equal the radius $5\sqrt{2}$ .

Set distance equal to the radius and compare with answer choices

Set the distance equal to the radius and solve for $b$ :

\frac{|14 + b|}{\sqrt{17}} = 5\sqrt{2}.

Multiply both sides by $\sqrt{17}$ :

|14 + b| = 5\sqrt{2} \cdot \sqrt{17} = 5\sqrt{34}.

14 + b = 5\sqrt{34} \quad \text{or} \quad 14 + b = -5\sqrt{34}.

Solve each:

$b = -14 + 5\sqrt{34}$
$b = -14 - 5\sqrt{34}$

Thus the two possible tangent lines with slope $4$ are

y = 4x - 14 + 5\sqrt{34} \quad \text{and} \quad y = 4x - 14 - 5\sqrt{34}.

Among the answer choices, only

y = 4x - 14 + 5\sqrt{34}

appears, so that is the correct choice.

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