In the -plane, the circle with equation and the line with equation , where is a positive constant, intersect at exactly one point. What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 9 | Free SAT Prep | Aniko

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Question 9·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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In the $xy$ -plane, the circle with equation $x^2 + y^2 = 13$ and the line with equation $3y + 2x = k$ , where $k$ is a positive constant, intersect at exactly one point. What is the value of $k$ ?

Your answer

(Express the answer as an integer)

When a line and a circle are said to intersect at exactly one point, think "tangent" and use geometry instead of heavy algebra. Identify the circle’s center and radius, rewrite the line in standard form, then apply the point-to-line distance formula and set that distance equal to the circle’s radius. This approach quickly leads to an equation in the parameter (here, k) and is usually faster and less error-prone than solving the system and checking when the quadratic has exactly one solution (discriminant zero).

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Use the meaning of "exactly one point"

Think about what it means geometrically when a line and a circle intersect at exactly one point. How is that different from intersecting at two points or not at all?

Focus on the circle's center and radius

From $x^2 + y^2 = 13$ , identify the center and radius of the circle. How far from the origin is any point on this circle?

Relate distance from a point to a line

Rewrite $3y + 2x = k$ in the form $Ax + By + C = 0$ , and recall the formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ . What should that distance equal for the line to just touch the circle?

Use the condition that k is positive

When you solve the equation involving $|k|$ , you will get two possible values. Use the fact that $k$ is a positive constant to decide which one is valid.

Desmos Guide

Graph the circle

In Desmos, type x^2 + y^2 = 13 to graph the circle centered at the origin with radius $\sqrt{13}$ .

Graph the family of lines

Type 3y + 2x = k. Desmos will prompt you to create a slider for k; accept it so k becomes a movable parameter.

Adjust k to see tangency

Move the k slider and watch how the line moves relative to the circle. Find the value of k where the line just touches the circle at exactly one point (it should "kiss" the circle without crossing it).

Read the needed k-value

When the line is tangent (touching the circle at only one point), read the corresponding value of k from the slider; that is the constant the question is asking for.

Step-by-step Explanation

Interpret "intersect at exactly one point"

If a line and a circle in the $xy$ -plane intersect at exactly one point, the line is tangent to the circle. That means the distance from the center of the circle to the line is exactly equal to the radius of the circle.

Identify the circle's center and radius

The circle has equation $x^2 + y^2 = 13$ .

This is a circle centered at $(0,0)$ .
Its radius $r$ satisfies $r^2 = 13$ , so $r = \sqrt{13}$ .

Write the line in standard form and use the distance formula

Rewrite the line $3y + 2x = k$ in standard form $Ax + By + C = 0$ :

Rearranging: $2x + 3y - k = 0$ .
So $A = 2$ , $B = 3$ , and $C = -k$ .

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

\text{distance} = \frac{|A\cdot 0 + B\cdot 0 + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}.

Here, that distance is

\text{distance} = \frac{|{-k}|}{\sqrt{2^2 + 3^2}} = \frac{|k|}{\sqrt{13}}.

Set distance equal to the radius and solve for k

For tangency, the distance from the center to the line must equal the radius $\sqrt{13}$ :

\frac{|k|}{\sqrt{13}} = \sqrt{13}.

Multiply both sides by $\sqrt{13}$ :

|k| = 13.

Since the problem states that $k$ is a positive constant, we choose $k = 13$ as the value that makes the line tangent to the circle.

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