The equation of the circle whose centre lies on the line $x-4 y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

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Aicharya · Accepted Answer

Let the equation of the circle be $x^2+y^2+2 g x+2 f y+c=0$ $\because$ its centre lies on $\mathrm{x}-4 \mathrm{y}=1$ $\Rightarrow-\mathrm{g}-4(-\mathrm{f})=1$ $\because$ it passes through $(3,7)$ $\Rightarrow 3^2+7^2+2 \mathrm{~g} 	imes 3+2 \mathrm{f} 	imes 7+\mathrm{c}=0$ Also it passes through $(5,5)$ $\Rightarrow 5^2+5^2+2 g 	imes 5+2 f 	imes 5+c=0$ from (iii) - (ii) $4 g-4 f=8 \Rightarrow g-f=2$ $\begin{aligned} & \Rightarrow \mathrm{f}=1 \ & \Rightarrow \mathrm{g}=3 \ & \Rightarrow \mathrm{c}=-90 \end{aligned}$ So, the equation of required circle is $x^2+y^2+6 x+2 y-90=0$

The equation of the circle whose centre lies on the line $x-4 y=1$ and which passes through the points $(3…

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