IIT-JEE 1978Question 1
Find the equation of the circle whose radius is $5$ and which touches the circle $x^{2}+y^{2}-2x-4y-20=0$ at the point $(5,5)$.
Circle
Answer$x^{2}+y^{2}-18x-16y+120=0$
Solution
Concept. The given circle also has radius $5$. A different circle of radius $5$ touching it at $(5,5)$ must touch externally, with centre the reflection of $(1,2)$ in $(5,5)$. Formulas used. (1) $C_2=2T-C_1$ (contact point $T$ is the midpoint of the centres).Step 1 - Given circle. Centre $(1,2)$, $r=\sqrt{1+4+20}=5$. $(5,5)$: $25+25-10-20-20=0$, on it. Step 2 - Use (1). $C_2=(10-1,\,10-2)=(9,8)$.Step 3 - Circle. $(x-9)^2+(y-8)^2=25\Rightarrow x^2+y^2-18x-16y+120=0$. Answer: $x^{2}+y^{2}-18x-16y+120=0$