Let the mirror image of a circle c_1:x^2+y^2-2x-6y+=0 in the line y=x+1 be c_2:5x^2+5y^2+1…

Question

Let the mirror image of a circle $c_1:x^2+y^2-2x-6y+\alpha=0$ in the line $y=x+1$ be $c_2:5x^2+5y^2+10gx+10fy+38=0$. If $r$ is the radius of circle $c_2$, then $\alpha+6r^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

$c_1$ has centre $(1,3)$, radius$^2=10-\alpha$. Reflecting $(1,3)$ in $x-y+1=0$: with $d=\frac{1-3+1}{2}=-\frac12$,
$$x'=1-2(1)(-\tfrac12)=2,\qquad y'=3-2(-1)(-\tfrac12)=2.$$

$c_2:x^2+y^2+2gx+2fy+\frac{38}5=0$, centre $(-g,-f)=(2,2)\Rightarrow g=f=-2$.

$$r^2=g^2+f^2-\frac{38}5=8-\frac{38}5=\frac25.$$

Since reflection preserves radius: $\frac25=10-\alpha\Rightarrow\alpha=\frac{48}5$.

$$\alpha+6r^2=\frac{48}5+\frac{12}5=12.$$

Solutions · 1