Question 124·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

In center-radius form $(x-h)^2+(y-k)^2=r^2$, the center is $(h,k)$. Since $y+2=y-(-2)$, circle A’s center is $(4,-2)$, not $(4,2)$.

A point lies on a circle when its coordinates make the circle’s equation true. So put circle A’s center into circle B’s equation, using $x=4$ and $y=-2$: $(4-(m+1))^2+(-2-(2m-1))^2=40$. Simplify inside the parentheses: $(3-m)^2+(-2m-1)^2=40$. Keep the negative $y$-coordinate when you substitute.

Type $(3-m)^2+(-2m-1)^2\sim40\{m>0\}$. The $\sim$ starts a regression, which finds a value that makes the two sides equal. Every listed value is positive, so the $m>0$ restriction tells Desmos which solution to find. Under PARAMETERS, it shows $m\approx2.658$.

Type the four choices on separate Desmos lines. They print about $2.658$, $1.329$, $2.258$, and $2.215$, in that order. The matching value is $\frac{1+\sqrt{151}}{5}$, so that is a possible value of $m$. Choice A.

Expand both squares in the point-on-circle equation from above: $(m^2-6m+9)+(4m^2+4m+1)=40$. The $4m$ is positive: the middle term of $(-2m-1)^2$ multiplies two negative terms.

Combine like terms: $5m^2-2m+10=40$. Subtract $40$ from both sides: $5m^2-2m-30=0$. The squared terms add to $5m^2$, which will determine the denominator in the formula.

For a quadratic equation $am^2+bm+c=0$, the formula is $m=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=5$, $b=-2$, and $c=-30$. Substitute them, keeping $b$ negative: $m=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-30)}}{2(5)}$.

Calculate under the square root: $m=\frac{2\pm\sqrt{604}}{10}$. Since $604=4\cdot151$, take $\sqrt{4}=2$ outside the root: $m=\frac{2\pm2\sqrt{151}}{10}$. Divide both numerator terms and the denominator by $2$: $m=\frac{1\pm\sqrt{151}}{5}$.

Because $\sqrt{151}>1$, the solution with the minus sign is negative. The listed values are positive, so keep the plus-sign solution: $m=\frac{1+\sqrt{151}}{5}$. Choice A.