Circles
Orthogonal circle locus and distance condition
MJAT_TS3_P2
Grade 12
Question:
A circle passes through the point $(3,4)$ and cuts the circle $x^2+y^2=a^2$ orthogonally. The locus of its centre is a straight line. If the distance of this straight line from the origin is $3$, then $a^2$ equals:
Step-by-Step Solution
Key Concept: Circle $x^2+y^2+2gx+2fy+c=0$ passes through $(3,4)$: $6g+8f+c=-25$. Orthogonal to $x^2+y^2=a^2$: $2(g)(0)+2(f)(0)=c-a^2\Rightarrow c=a^2$. Locus of centre $(-g,-f)$: substitute $c=a^2$ into passing condition: $6g+8f=-25-a^2$, i.e., $6x+8y=25+a^2$.
$a^2=\mathbf{5}$.
Correct Answer: 5