The graphs of the circle and the line intersect in the -plane. For which of the following values of do the two graphs intersect at exactly one point?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 35 | Free SAT Prep | Aniko

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Question 35·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The graphs of the circle $x^2 + y^2 = 25$ and the line $y = kx + 5$ intersect in the $xy$ -plane. For which of the following values of $k$ do the two graphs intersect at exactly one point?

When a question asks for a line and a circle to intersect at exactly one point, translate that condition into “the line is tangent to the circle.” Quickly identify the circle’s center and radius, then either (1) use the point-to-line distance formula and set the distance from the center to the line equal to the radius, or (2) substitute the line equation into the circle’s equation and require that the resulting quadratic have exactly one solution (discriminant zero). Checking the given answer choices with this condition lets you solve efficiently without unnecessary algebra.

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Relate “exactly one point” to a special position of the line

Think about how a line can meet a circle: in how many points can they intersect, and what do we call it when they touch in only one point?

Use the circle’s center and radius

The circle $x^2 + y^2 = 25$ has center at $(0,0)$ and radius $5$ . For which positions of the line $y = kx + 5$ will the closest point on the line to $(0,0)$ be exactly $5$ units away?

Distance from a point to a line

Rewrite the line as $kx - y + 5 = 0$ . Use the distance formula from a point to a line:

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

with $(x_0, y_0) = (0,0)$ and $A = k$ , $B = -1$ , $C = 5$ , and then set that distance equal to the radius $5$ .

Desmos Guide

Graph the circle

In Desmos, enter the equation x^2 + y^2 = 25 to graph the circle with radius 5 centered at the origin.

Graph the family of lines with a slider for k

Enter y = kx + 5. Desmos will create a slider for k, letting you vary the slope while keeping the y-intercept at 5.

Test each answer choice for k

Move the k slider (or type in the values) to check $k = -4$ , $k = -2$ , $k = 0$ , and $k = 3$ . For each value, look at how many points the line and circle share: two intersection points means it cuts through the circle, while exactly one visible intersection means the line is tangent.

Identify the correct k from the graph

Compare the four cases and note which value of $k$ makes the line just touch the circle at a single point without crossing through it. That $k$ corresponds to the correct answer choice.

Step-by-step Explanation

Interpret “exactly one point” geometrically

A line and a circle in the plane can intersect in 0, 1, or 2 points.

If they intersect in two points, the line is a secant (it cuts through the circle).
If they intersect in one point, the line is tangent to the circle.

So we are looking for the value of $k$ that makes the line $y = kx + 5$ tangent to the circle $x^2 + y^2 = 25$ .

Use the circle’s center and radius

The circle $x^2 + y^2 = 25$ has

center at $(0,0)$ , and
radius $5$ (since $25 = 5^2$ ).

For a line to be tangent to a circle, the distance from the circle’s center to the line must be equal to the radius. Here, that distance must be $5$ .

Write the line in standard form and apply the distance formula

Rewrite the line $y = kx + 5$ in the standard form $Ax + By + C = 0$ :

$y = kx + 5 \Rightarrow kx - y + 5 = 0$ , so $A = k$ , $B = -1$ , $C = 5$ .

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

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

Here, the point is the center $(0,0)$ , so

d = \frac{|k\cdot 0 + (-1)\cdot 0 + 5|}{\sqrt{k^2 + 1}} = \frac{5}{\sqrt{k^2 + 1}}.

For tangency, this distance must equal the radius $5$ :

Solve the distance equation for k and pick the answer

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

\frac{5}{\sqrt{k^2 + 1}} = 5 \\[4pt] \frac{1}{\sqrt{k^2 + 1}} = 1 \\[4pt] \sqrt{k^2 + 1} = 1 \\[4pt] k^2 + 1 = 1 \\[4pt] k^2 = 0.

Thus $k = 0$ .

So the line $y = 0\cdot x + 5$ , which is $y = 5$ , is tangent to the circle and intersects it at exactly one point. The correct answer choice is C) 0.

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