Find the equation of circle having normals \((x-1)(y-2)=0\) and a tangent \(3 x+4 y=6\) ?
Find the equation of circle having normals \((x-1)(y-2)=0\) and a tangent \(3 x+4 y=6\) ?
\((x-1)^2+(y-2)^2=1\)
\((x-2)^2+(y-1)^2=1\)
\((x+1)^2+(y+2)^2=1\)
\((x+2)^2+(y+1)^2=1\)
Solution
The equation of normals to the circle are \(x-1=0\) and \(y-2=0\), so centre of the circle is \((1,2)\) and since \(3 x+4 y=6\) is the tangent to the circle so radius \(r=\frac{3+8-6}{\sqrt{3^2+4^2}}=1\)
\(\therefore\) Equation of required circle is
\((x-1)^2+(y-2)^2=1\)
Hence, option (a) is correct.