let the equation of circle be-
\(x ^2
+y ^
2
+2gx+2fy+c=0\)
centre of circle = (-g,-f)
radius of circle = \( \sqrt{f^2+g^2-c } = 5(given)
\)
distance between the centre of circle and the line 3x+4y-11=0 is radius of the circle, as it touches the circle, i.e.5
therefore,
\( {|-3g-4f-11|\over \sqrt{3^2+4^2} }=5\)
→ -3g-4f-11=25
→ -3g-4f=36 ....(1)
now slope of normal to the circle at (1,2)
\(m = {2+f \over 1+g}\)
slope of the given line, \(m' = {-3 \over 4}\)
therefore mm'=-1(normal and tangent to the circle at a point are always perpendicular)
→\( {2+f \over 1+g}={4\over 3}\)
→ 6+3f=4+4g
→ 4g-3f=2 ....(2)
multiply equation (1) and (2) by 4 and 3 respectively,
→ -12g-16f=144 ....(3)
→ 12g-9f=6 ....(4)
add equation (3) and (4) we get,
→ -25f=150
→ f=\(
{150\over -25}=-6\)
substituting value of f in equation (1).
→ -3g+24=36
→ \(
g= {36-24\over -3}=-4\)
thus the centre of circle is (4,6)
now, put the values of g and f in radius of circle equation \( \sqrt{f^2+g^2-c } = 5(given)
\)
→ \( \sqrt{16+36-c } = 5
\)
sqaring both sides we get,
52-c=25
→ c= 52-25= 27
hence the equation of circle is
\(x^2+y^2-8x-12y+27=0\)