Question 752·Hard·Math

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In the $xy$ -plane, there is exactly one circle that satisfies all of the following conditions:

Its center lies on the $x$ -axis.
It is tangent to the line $y = x + 1$ .
It passes through the point $(4, 5)$ .

Which of the following is an equation of this circle?

The key to tangency is finding where the line touches the circle. If a given point lies on both, a radius to that point is perpendicular to the line; follow that radius to the required center axis. Then use the point to find the radius squared. In the finished equation, watch the sign of the center’s coordinate.

Hints

Hint 1
A tangent line meets a circle at exactly one point. Check whether the point given on the circle also satisfies the line’s equation. If it does, what does that tell you about where the line touches?
Hint 2
At the touching point, a radius is perpendicular to the tangent line. This line rises one unit for each unit right, so its perpendicular falls one unit for each unit right. Follow it to the $x$ -axis.
Hint 3
The squared radius is the horizontal change squared plus the vertical change squared. Use the center and the given point, then put that result on the right side of the circle equation.

Step-by-step

Approach 1: Find the center from the tangent point

Step 1
Identify where the line touches
Name the line’s output $f(x)=x+1$ . Type $f(x)=x+1$ and $f(4)$ in Desmos; it shows $5$ . So $(4,5)$ lies on the line as well as on the circle. A tangent line meets a circle only once, so this is the touching point.
Step 2
Follow the radius to the center
The radius to a touching point is perpendicular to the tangent line. The line’s slope, its rise per unit right, is $1$ , so the perpendicular’s slope is $-1$ . From $(4,5)$ , go $5$ down to reach the $x$ -axis. That also means going $5$ right, putting the center at $(4+5,0)=(9,0)$ .
Step 3
Find the squared radius
From the center $(9,0)$ to $(4,5)$ , the horizontal and vertical changes each have size $5$ . By the Pythagorean theorem, square and add them: $r^2=5^2+5^2$ . Type $5^2+5^2$ in Desmos; it shows $50$ . That’s $r^2$ , the number the equation needs.
Step 4
Write the circle equation
A circle with center $(h,k)$ and squared radius $r^2$ has standard form $(x-h)^2+(y-k)^2=r^2$ . Use center $(9,0)$ and $r^2=50$ : $(x-9)^2+y^2=50$ . Use $x-9$ , not $x+9$ ; the latter would put the center at $-9$ . Choice D.

Approach 2: Test the given point against the choices

Step 1
Reject two centers using the point
For a point to be on a circle, its coordinates must make the equation’s two sides equal. In Desmos, set $x_1=4$ and $y_1=5$ , then evaluate the left sides with $x+9$ and $x-5$ . Desmos shows $194$ and $26$ , respectively. Both equations have $50$ on the right, so neither contains $(4,5)$ .
Step 2
Compare the remaining right sides
Add $(x_1-9)^2+y_1^2$ in Desmos; it shows $50$ . The two remaining equations share this left side, but one sets it equal to $32$ and the other to $50$ . Only $(x-9)^2+y^2=50$ contains the given point. Because the problem says exactly one circle satisfies all the conditions, you can select it without another calculation. Choice D.

Question 752·Hard·Math

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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