Q80 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
The equation of the circle whose entre lies on the line $x-4 y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
- A$x^2+y^2+6 x-2 y+90=0$
- B$x^2+y^2-6 x-2 y-25=0$
- C$x^2+y^2-6 x+2 y-30=0$
- Correct answerD$x^2+y^2+6 x+2 y-90=0$
Solution
Let $(h, k) b$ the centre of the circle. It lies on the line $x-4 y=1$
$$
\begin{aligned}
& \Rightarrow \mathrm{h}=1+4 \mathrm{k} \\
& \therefore \text { centre } \equiv(4 \mathrm{k}+1, \mathrm{k})
\end{aligned}
$$
Circle passes through points $(3,7)$ and $(5,5)$
$$
\begin{aligned}
& \therefore(4 \mathrm{k}+1-5)^2+(\mathrm{k}-5)^2=(4 \mathrm{k}-2)^2+(\mathrm{k}-7)^2 \\
& \therefore 16 \mathrm{k}^2+16-32 \mathrm{k}+\mathrm{k}^2+25-10 \mathrm{k}=16 \mathrm{k}^2+4-16 \mathrm{k}+\mathrm{k}^2+49 \\
& -14 \mathrm{k}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore-42 \mathrm{k}+41=-30 \mathrm{k}+53 \Rightarrow 12 \mathrm{k}=-12 \Rightarrow \mathrm{k}=-1 \\
& \therefore \text { centre } \equiv(-3,-1) \\
& \therefore \text { Radius }=\sqrt{(-3-5)^2+(-1-5)^2}=10
\end{aligned}
$$
Hence equation of required circle is
$$
(x+3)^2+(y+1)^2=(10)^2 \text { i.e. } x^2+y^2+6 x+2 y-90=0
$$
Hence, the correct answer is option: 'D'.