Question 88·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A circle has equation $x^2 + y^2 - 6x + 8y = m$ , where $m$ is a constant. The line $3x + 4y = 24$ is tangent to the circle.

Which choice gives the value of $m$ ?

An expanded circle equation hides its center and radius. Complete the square to find them, then use the fact that a tangent line sits exactly one radius from the center. Desmos can calculate that distance and solve for the constant. The main trap is stopping at the radius squared instead of finding the constant the question asks for.

Hints

Hint 1
Completing the square turns the expanded equation into $(x-h)^2+(y-k)^2=r^2$ . Add the square of half the $x$ coefficient to both sides, then do the same for the $y$ coefficient. What center does that reveal?
Hint 2
A tangent touches a circle at one point. The radius to that point is perpendicular to the line, so the shortest distance from the center to the line equals the radius.
Hint 3
The right side of the completed circle equation is $r^2$ , not $m$ alone. Once you know the radius, what must you subtract to isolate $m$ ?

Step-by-step

Approach 1: Find the center-to-line distance
Step 1
Complete the $x$ square
Half of $-6$ is $-3$ , and $(-3)^2=9$ . Add $9$ to both sides so the $x$ terms make a square:
$x^2-6x+9+y^2+8y=m+9.$
Rewrite those terms:
$(x-3)^2+y^2+8y=m+9.$
Step 2
Complete the $y$ square
Half of $8$ is $4$ , so add $4^2=16$ to both sides:
$(x-3)^2+y^2+8y+16=m+9+16.$
Rewrite the circle equation in center-radius form:
$(x-3)^2+(y+4)^2=m+25.$
So its center is $(3,-4)$ , and its radius squared is $m+25$ . The plus sign in $y+4$ means the center’s $y$ -coordinate is $-4$ .
Step 3
Find the radius from the tangent line
A tangent touches the circle once, so the shortest distance from its center to the line is the radius. For $Ax+By=C$ , the distance from $(h,k)$ is $\frac{|Ah+Bk-C|}{\sqrt{A^2+B^2}}$ . Type $\frac{|3(3)+4(-4)-24|}{\sqrt{3^2+4^2}}$ in Desmos. It shows $6.2$ ; the fraction button shows $r=\frac{31}{5}$ .
Step 4
Solve for the circle’s constant
The completed equation says $m+25=r^2$ . Type $m+25\sim\left(\frac{31}{5}\right)^2$ so Desmos solves for $m$ . It shows $m=13.44$ under PARAMETERS. Type $m$ on the next line and use its fraction button to see $\frac{336}{25}$ . The squared radius includes the $25$ added while completing the squares, so $m=\frac{336}{25}$ . Choice A.

Approach 2: Use the one-point tangency condition
Step 1
Write the line as an expression for $y$
A point on the line satisfies $3x+4y=24$ . Subtract $3x$ :
$4y=24-3x.$
Divide by $4$ :
$y=6-\frac{3x}{4}.$
Step 2
Substitute the line into the circle
At a meeting point, both equations hold. Put $6-\frac{3x}{4}$ in for $y$ in the circle equation:
$x^2+\left(6-\frac{3x}{4}\right)^2-6x+8\left(6-\frac{3x}{4}\right)=m.$
Step 3
Put the result in quadratic form
Expand the squared term and distribute the $8$ :
$x^2+36-9x+\frac{9x^2}{16}-6x+48-6x=m.$
Combine like terms:
$\frac{25}{16}x^2-21x+84=m.$
Subtract $m$ :
$\frac{25}{16}x^2-21x+84-m=0.$
Step 4
Make the quadratic have exactly one solution
A tangent meets the circle once, so this quadratic has one real solution. Its discriminant, $b^2-4ac$ for $ax^2+bx+c=0$ , must be $0$ : that makes the plus and minus solutions of the quadratic formula coincide. Here $a=\frac{25}{16}$ , $b=-21$ , and $c=84-m$ , so:
$(-21)^2-4\left(\frac{25}{16}\right)(84-m)=0.$
Step 5
Solve the one-solution condition
Type the discriminant equation with $\sim$ so Desmos solves for $m$ . PARAMETERS shows $m=13.44$ . Type $m$ below it and use the fraction button: the constant is $\frac{336}{25}$ . Choice A.

Question 88·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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