A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\).…
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is.........
\(4 x^2+4 y^2-40 x-16 y+67=0\)
\(3 x^2+3 y^2-40 x-16 y+67=0\)
\(2 x^2+2 y^2-40 x-16 y+67=0\)
\(x^2+y^2-10 x-4 y+67=0\)
Solution
The centre of the circle \(x^2+y^2-3 x-4 y-1=0\) is \(C(3 / 2,2)\), now as it is given that the required circle passes through the point \(C(3 / 2,2)\) and having centre \((5,2)\), so radius of required circle is
\(r=\sqrt{\left(5-\frac{3}{2}\right)^2+(2-2)^2}=\frac{7}{2}\)
\(\therefore\) Equation of required circle is
\(\begin{aligned}
(x-5)^2+(y-2)^2 & =\left(\frac{7}{2}\right)^2 \\
\Rightarrow 4 x^2+4 y^2-40 x-16 y+116 & =49 \\
\Rightarrow 4 x^2+4 y^2-40 x-16 y+67 & =0
\end{aligned}\)
Hence, option (a) is correct.