Question

If the equation of tangent to a circle at point (3, 5) is $ 2x-y-1=0 $ and its centre lies on $ x+y=5, $ then the equation of circle is

• A. $ {{x}^{2}}+{{y}^{2}}+6x-16y+28=0 $
• B. $ {{x}^{2}}+{{y}^{2}}-6x-16y+28=0 $
• C. $ {{x}^{2}}+{{y}^{2}}+6x+6y+28=0 $
• D. $ {{x}^{2}}+{{y}^{2}}-6x-6y-28=0 $

Answer 1

Answer: (A)

Equation of line which is perpendicular to the tangent
$ 2x-y-1=0 $ is $ x+2y+k=0 $ .
Equation of tangent passes through the point (3,5).
$ \therefore $ $ 3+2\times 5+k=0 $
$ \Rightarrow $ $ k=-13 $
$ \therefore $ Centre of circle will lie on the perpendicular line
$ x+2y=13 $ .
Also, centre of circle lies on the line $ x+y=5 $ .
So, on solving these two equations, we get
$ x=-3,y=8 $
Centre $ =(-3,8) $
and radius $ =\sqrt{{{(3+3)}^{2}}+{{(5-8)}^{2}}} $
$ =\sqrt{{{6}^{2}}+{{3}^{2}}} $
$ =\sqrt{45} $
$ \therefore $ Equation of circle is $ {{(x+3)}^{2}}+{{(y-8)}^{2}}={{(\sqrt{45})}^{2}} $
$ \Rightarrow $ $ {{x}^{2}}+9+6x+{{y}^{2}}+64-16y=45 $
$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}+6x-16y+28=0 $