Circle: If a circle passes through the point (a,b) and cuts the circle… | IIT-JEE 1988

IIT-JEE 1988Question 1

If a circle passes through the point $(a,b)$ and cuts the circle $x^{2}+y^{2}=k^{2}$ orthogonally, then the equation of the locus of its centre is

Circle

$2ax+2by-\left(a^{2}+b^{2}+k^{2}\right)=0$✓ Correct answer$2ax+2by-\left(a^{2}-b^{2}+k^{2}\right)=0$$x^{2}+y^{2}-3ax+4by+\left(a^{2}+b^{2}-k^{2}\right)=0$$x^{2}+y^{2}-2ax-3by+\left(a^{2}-b^{2}-k^{2}\right)=0$.

Solution

Concept. General circle, orthogonality, pass through $(a,b)$, then write the centre as $(x,y)$. Formulas used. (1) Orthogonal: $2g_1g_2+2f_1f_2=c_1+c_2$.Step 1 - Use (1). With $x^2+y^2-k^2=0$: $0=c-k^2\Rightarrow c=k^2$. Step 2 - Through $(a,b)$. $a^2+b^2+2ga+2fb+k^2=0$. Step 3 - Centre $(x,y)=(-g,-f)$. $$2ax+2by-(a^2+b^2+k^2)=0.$$ Answer: A

Circle - JEE Mathematics question (IIT-JEE 1988)

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