Find the equation of the circle the end points of whose diameter are the centres of the circles…

Find the equation of the circle the end points of whose diameter are the centres of the circles \(\mathrm{x}^2+\mathrm{y}^2+16 \mathrm{x}-14 \mathrm{y}-1=0 ~\&~ \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+10 \mathrm{y}-2=0\)
  1. 20
  2. 30
  3. 40
  4. 61

Solution

Centre of first circle
\(\begin{aligned}
& x^2+y^2+16 x-14 y-1=0 . \\
& \text { is }(-g,-f) \\
& \Rightarrow(-8,7) \rightarrow P
\end{aligned}\)
For II Circle:
Centre \(\rightarrow(2,-5) \rightarrow Q\)
Equation of circle with \(P \& Q\) as diametric end points:
\(\begin{aligned}
& \text {Centre } \rightarrow \quad\left(\frac{-8+2}{2}, \frac{7-5}{2}\right) \\
& \Rightarrow(-3, \quad 1) . \\
& \text {radius }=\frac{1}{2} \sqrt{(2+8)^2+(-5-7)^2} \\
& =\frac{1}{2} \sqrt{100+144}=\frac{\sqrt{244}}{2} \\
& \Rightarrow(x+3)^2+(y-1)^2=\frac{244}{4} \\
& \Rightarrow(x+3)^2+(y-1)^2=61
\end{aligned}\)

Asked in: MHT CET 2020 (20 Oct Shift 1)

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