In the -plane, a circle is tangent to the -axis at the point and passes through the point . What is the radius of the circle?

Solution available with step-by-step explanation

SAT Circles Question 34 | Free SAT Prep | Aniko

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Question 34·Hard·Circles

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In the $xy$ -plane, a circle is tangent to the $y$ -axis at the point $(0,2)$ and passes through the point $(3,6)$ . What is the radius of the circle?

For circle problems involving tangents on the SAT, quickly sketch the situation and use the fact that a radius to a point of tangency is perpendicular to the tangent line. This often pins down the center's coordinates in terms of the radius. Then apply the distance formula from the center to a given point on the circle, set that distance equal to the radius, and solve the resulting equation carefully, watching your algebra when you expand and simplify.

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Think about the picture

Make a quick sketch: draw the y-axis, mark the point (0,2), and draw a circle touching the y-axis at that point and passing through (3,6). Where would the center of that circle lie relative to (0,2)?

Use the tangent property

For a circle, the radius to a point of tangency is perpendicular to the tangent line. Since the y-axis is vertical, what direction must the radius from the center to (0,2) go, and what does that say about the center's coordinates?

Introduce a variable for the radius

Call the radius $r$ , and let the center be $(r,2)$ . Use the distance formula between $(r,2)$ and $(3,6)$ and set that distance equal to $r$ to form an equation.

Solve the equation carefully

After you write $(3 - r)^2 + 4^2 = r^2$ , expand and simplify step by step to solve for $r$ . Make sure you do not forget to square the radius on the right-hand side.

Desmos Guide

Set up the radius equation in Desmos

Let $x$ represent the radius. In Desmos, enter the expression as a function, for example: y = (3 - x)^2 + (6 - 2)^2 - x^2. This comes from rearranging $(3 - x)^2 + (6 - 2)^2 = x^2$ so that everything is on one side.

Find the positive solution

Look at the graph of this function and identify where it crosses the x-axis (the x-intercepts). The positive x-intercept is the value of the radius that satisfies the equation.

Step-by-step Explanation

Use the tangency to locate the center

A circle tangent to the y-axis at (0,2) touches the vertical line $x=0$ there.

The radius to a tangent point is always perpendicular to the tangent line.
The y-axis is vertical, so the radius at (0,2) must be horizontal.

That means the center has the same y-coordinate as the tangent point and is some horizontal distance $r$ away. So we can write the center as

(r, 2)

where $r$ is the radius (the horizontal distance from $x=0$ to the center).

Write an equation using the distance formula

The distance from the center $(r,2)$ to any point on the circle equals the radius $r$ .

The circle passes through $(3,6)$ , so the distance from $(r,2)$ to $(3,6)$ must be $r$ .

Using the distance formula:

\sqrt{(3 - r)^2 + (6 - 2)^2} = r

Square both sides to remove the square root:

(3 - r)^2 + (6 - 2)^2 = r^2

Substitute $6 - 2 = 4$ and $4^2 = 16$ :

(3 - r)^2 + 16 = r^2

Now expand $(3 - r)^2$ :

(3 - r)^2 = 9 - 6r + r^2

So the equation becomes

9 - 6r + r^2 + 16 = r^2.

Solve for the radius

Start with

9 - 6r + r^2 + 16 = r^2.

Combine like terms on the left:

25 - 6r + r^2 = r^2.

Subtract $r^2$ from both sides:

25 - 6r = 0.

Solve for $r$ :

\begin{aligned} -6r &= -25 \\ r &= \frac{25}{6} \end{aligned}

Because $r$ is a radius, it must be positive, so the radius of the circle is

\frac{25}{6}.

This corresponds to choice C.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation