The equation of the çircle which has its centre at the point $(3,4)$ and touches the line $5 x+12 y-11=0$ is

The equation of the çircle which has its centre at the point $(3,4)$ and touches the line $5 x+12 y-11=0$ is
  1. $x^2+y^2-6 x-8 y+9=0$
  2. $x^2+y^2-6 x-8 y+25=0$
  3. $x^2+y^2-6 x-8 y-9=0$
  4. $x^2+y^2-6 x-8 y-25=0$

Solution

$\begin{aligned} & \text { Radius }=\text { Distance of a point }(3,4) \text { form } \\ & 5 x+12 y-11=0 \\ & =\left|\frac{5(3)+12(4)-11}{\sqrt{25+144}}\right| \\ & =\left|\frac{15+48-11}{\sqrt{169}}\right| \\ & =\frac{52}{13} \\ & =4 \end{aligned}$ $\therefore \quad$ Required equation is $\begin{aligned} & (x-3)^2+(y-4)^2=(4)^2 \\ & x^2+y^2-6 x-8 y+9=0 \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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