A circle \(S\) of radius 2 units lies in the first quadrant and touches both the coordinate axes. The…
- \(x^2+y^2-12 x-10 y+12=0\)
- \(x^2+y^2-12 x-10 y-20=0\)
- \(x^2+y^2-12 x-10 y+25=0\)
- \(x^2+y^2-12 x-10 y+52=0\)
Solution

Centre of given circle \(\left(C_1\right)=(2,2)\) radius \(=2\) units Centre of required circle \(\left(C_2\right)=(6,5)\) $\begin{aligned} C_{1} C_{2} & =r+2 \quad \text{(from figure)} \\ \therefore \quad r & =r+2 \end{aligned}$ Therefore, the required equation of the circle having centre at $(6,5)$ and $r=3$ is $\begin{aligned} (x-6)^{2}+(y-5)^{2} & =3^{2} \\ x^{2}+y^{2}-12x-10y+52 & =0 \end{aligned}$ Therefore, the answer is (d).
Asked in: AP EAMCET 2019 (23 Apr Shift 1)