The equation of the circle, which touches the line $y=5$ and passes through $(-1,2)$ and $(1,2)$ is

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Aicharya · Accepted Answer

The centre of the circle is on the perpendicular bisector of the line joining $(-1,2)$ and $(1,2)$, which is the $y$-axis. The ordinate of the centre is given by $\begin{array}{l} (5-y)^{2}=1+(y-2)^{2} \\ \Rightarrow y=\frac{10}{3} \end{array}$ Hence, eq. of the circle is: $\begin{array}{l} x^{2}+\left(y-\frac{10}{3}\right)^{2}=\left(\frac{5}{3}\right)^{2} \\ \Rightarrow 9 x^{2}+9 y^{2}-60 y+75=0 \end{array}$

The equation of the circle, which touches the line $y=5$ and passes through $(-1,2)$ and $(1,2)$ is

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