In the -plane, the line with equation is tangent to the circle If , what is the value of ?

Solution available with step-by-step explanation

SAT Circles Question 12 | Free SAT Prep | Aniko

00:00

Question 12·Hard·Circles

0 / 50

In the $xy$ -plane, the line with equation $y = mx + 3$ is tangent to the circle $(x - 2)^2 + (y + 1)^2 = 10$ . If $m > 0$ , what is the value of $m$ ?

Your answer

(Express the answer as an integer)

For line–circle tangency problems, first rewrite the circle in standard form to identify the center and radius quickly. Then either (1) use the point-to-line distance formula and set that distance equal to the radius, or (2) substitute the line equation into the circle and set the quadratic’s discriminant to zero; in both methods, you end with an equation in the slope parameter. Simplify carefully to a quadratic, solve using factoring or the quadratic formula, and finally apply any given conditions (such as $m>0$ ) to select the correct solution. This approach is systematic and avoids messy guessing about the geometry of the tangent point.

Hints

Click me!

Use the circle’s standard form

Rewrite the circle $(x - 2)^2 + (y + 1)^2 = 10$ in the form $(x - h)^2 + (y - k)^2 = r^2$ and identify the center $(h,k)$ and the radius $r$ .

Relate tangency to distance

For a line tangent to a circle, how many points of intersection do they have? How can you express this situation in terms of the distance from the circle’s center to the line?

Apply the point-to-line distance formula

Rewrite $y = mx + 3$ in the form $Ax + By + C = 0$ , then use the distance formula $\dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ with the center of the circle as $(x_0,y_0)$ .

Solve the resulting equation for m

Set the distance from the center to the line equal to the circle’s radius, square both sides to remove square roots, simplify to a quadratic equation in $m$ , and then solve that quadratic. Finally, use the condition $m > 0$ to decide which solution to keep.

Desmos Guide

Translate the algebraic condition into two functions

From the distance setup, you get the equation $4(m+2)^2 = 10(m^2+1)$ . In Desmos, treat $x$ as $m$ and enter two functions:

y = 4(x + 2)^2
y = 10(x^2 + 1)

Find the intersection points

Use Desmos’s intersection tool to find where the two graphs intersect. There will be two intersection points; their $x$ -coordinates are the two solutions to the equation.

Choose the correct slope

Look at the $x$ -coordinates of the intersection points you found. The positive $x$ -value corresponds to the required slope $m$ in the original problem.

Step-by-step Explanation

Identify the circle’s center and radius

The equation of the circle is

(x - 2)^2 + (y + 1)^2 = 10.

This is in the form $(x - h)^2 + (y - k)^2 = r^2$ , so:

Center: $(h,k) = (2,-1)$
Radius: $r = \sqrt{10}$ because $r^2 = 10$ .

Write the distance from the center to the line

The line has equation $y = mx + 3$ . Rewrite it in the form $Ax + By + C = 0$ :

y - mx - 3 = 0,

so $A = -m$ , $B = 1$ , and $C = -3$ .

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}}.

Here the point is the center $(2,-1)$ , so

\text{distance} = \frac{|(-m)(2) + 1\cdot(-1) + (-3)|}{\sqrt{(-m)^2 + 1^2}}.

Simplify the numerator:

(-m)(2) + (-1) + (-3) = -2m - 4,

so the distance becomes

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

Set the distance equal to the radius and form an equation for m

Because the line is tangent to the circle, the distance from the center to the line must equal the radius $\sqrt{10}$ :

\frac{2|m + 2|}{\sqrt{m^2 + 1}} = \sqrt{10}.

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

\left(\frac{2|m + 2|}{\sqrt{m^2 + 1}}\right)^2 = (\sqrt{10})^2 \\[0.7em] \frac{4(m + 2)^2}{m^2 + 1} = 10.

Multiply both sides by $m^2 + 1$ :

4(m + 2)^2 = 10(m^2 + 1).

Expand both sides:

4(m^2 + 4m + 4) = 10m^2 + 10 \\[0.7em] 4m^2 + 16m + 16 = 10m^2 + 10.

Move all terms to one side:

0 = 10m^2 + 10 - 4m^2 - 16m - 16 \\[0.7em] 0 = 6m^2 - 16m - 6.

Divide by 2 to simplify:

3m^2 - 8m - 3 = 0.

Now solve this quadratic equation for $m$ . (We will do that in the next step.)

Solve the quadratic and use the condition m > 0

We have the quadratic equation

3m^2 - 8m - 3 = 0.

Use the quadratic formula $m = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 3$ , $b = -8$ , and $c = -3$ :

m = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2\cdot 3} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6}.

So the two possible values are

m = \frac{18}{6} = 3 \quad\text{or}\quad m = \frac{-2}{6} = -\frac{1}{3}.

The problem states $m > 0$ , so we choose the positive value.

Therefore, $m = 3$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation