If a circle of radius 3 passes through the point $(7,3)$ and -has its centre on the line $x-y-1=0$, then its equation among the following is

Question

Aicharya · Accepted Answer

Let $c(h, k)$ be the centre of the given circle.

$\because c(h, k)$ lies on the line $x-y-1=0$ $\Rightarrow \quad h-k-1=0$ $\Rightarrow \quad h=k+1 \quad \ldots$ (i) Now, $C P$ is the radius. $\begin{array}{lr}\Rightarrow & C P=3 \Rightarrow(C P)^2=9 \\ \Rightarrow & (h-7)^2+(k-3)^2=9 \\ \Rightarrow & (k-6)^2+(k-3)^2=9\end{array}$ $\begin{array}{ll}\Rightarrow & 2 k^2-18 k+36=0 \\ \Rightarrow & k^2-9 k+18=0\end{array}$ $\Rightarrow \quad(k-6)(k-3)=0 \Rightarrow k=\{6,3\}$ $\Rightarrow \quad h=\{6+1,3+1\}=\{7,4\}$ Thus, $(h, k) \equiv(7,6)$ or $(4,3)$. When $C \equiv(7,6)$ and $r=3$ Equation of circle $(x-7)^2+(y-6)^2=3^2$ $\Rightarrow \quad x^2+y^2-14 x-12 y+76=0$ $\therefore$ Option (d) is true.

If a circle of radius 3 passes through the point $(7,3)$ and -has its centre on the line $x-y-1=0$, then its…

Solution