The equation of a circle passing through the point $(2,8)$, touching the lines $4 x-3 y-24=0$ and $4 x+3 y-42=0$ and having the $x$ coordinate of its centre less than or equal to 8 is

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Aicharya · Accepted Answer

By the property of distance, $\begin{aligned} & \left|\frac{4 h-3 k-24}{5}\right|=\left|\frac{4 h+3 k-42}{5}\right| \ & =\sqrt{(h-2)^2+(k-8)^2} \ & \Rightarrow \quad 4 h-3 k-24= \pm(4 h+3 k-42) \ & \text {either } \quad 4 h-3 k-24=4 h+3 k-42 \ & \Rightarrow \quad 6 k=18 \Rightarrow k=3 \ & \text {or } \quad 4 h-3 k-24=-4 h-3 k+42 \ & \Rightarrow \quad 8 h=66 \ & \Rightarrow \quad h=\frac{66}{8}=\frac{33}{4} > 8 \end{aligned}$ Now, $(4 h-3 k-24)^2=25\left[(h-2)^2+(k-8)^2\right]$ By solving this, we get, $h=2$ $\therefore$ Centre is $(2,3)$. Now, required circle is $\begin{aligned} & (x-2)^2+(y-3)^2 =25 \ & \Rightarrow x^2+4-4 x+y^2+9-6 y =25 \ & \Rightarrow x^2+y^2-4 x-6 x =12 \end{aligned}$

The equation of a circle passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4…

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