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  • A circle C touches the line x = 2y at the point (2,1) and intersects the circle <img alt="\mathrm{C_1: x^2+y^2+2 y-5=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BC

A circle C touches the line x = 2y at the point (2,1) and intersects the circle \mathrm{C_1: x^2+y^2+2 y-5=0} at two points P and Q such that PQ is a diameter of \mathrm{C_1}. Then the diameter of C is.

Option: 1

4 \sqrt{15}


Option: 2

\sqrt{285}


Option: 3

15


Option: 4

7 \sqrt{5}


Answers (1)

best_answer

Let the equation of C be
\mathrm{C \equiv x^2+y^2+2 g x+2 f y+c=0}, then
Centre \mathrm{=(-g,-f)}
Radical axis of \mathrm{C ~\& ~C_1} is
\mathrm{ 2 g x+2(f-1) y+c+5=0 }
This passes through centre of \mathrm{ C_1 }, i.e., \mathrm{ (0,-1) } lies on radical axis.
\mathrm{ \Rightarrow-2(f-1)=-(c+5) \\ }
\mathrm{ \Rightarrow 2(f-1)=c+5 }       .....(i)

Equation of normal to circle C passing through (2,1) is
\mathrm{ y-1=-2(x-2) \quad }            \mathrm{ [\because x=2 y \text { is tangent at }(2,1)] \\ }
\mathrm{ \Rightarrow 2 x+y=5 }
\mathrm{ \because \quad } This passes through the centre of C i.e., (-g,-f)
\mathrm{ \therefore \quad-2 g-f=5 \\ }
\mathrm{ \Rightarrow \quad 2 g+f=-5 }         .....(ii)

\mathrm{\because \quad } The circle C passes through (2,1).
\mathrm{ \therefore \quad 4+1+4 g+2 f+c=0 \\ }
\mathrm{ \Rightarrow \quad 2 g+f=\frac{-(5+c)}{2}=-5 }  (from (ii))
\mathrm{ \Rightarrow 5+c=10 }
\mathrm{ \Rightarrow c=5 }
Now, from (i), we have \mathrm{2(f-1)=c+5}
\mathrm{ \Rightarrow 2(f-1)=10 }
\mathrm{ \Rightarrow f=6 }
\mathrm{ \Rightarrow 2 g=-5-f=-11 }
\mathrm{ \Rightarrow g=\frac{-11}{2} }

Thus, \mathrm{C \equiv x^2+y^2-11 x+12 y+5=0 }
\mathrm{\Rightarrow\left(x-\frac{11}{2}\right)^2+(y+6)^2=36-5+\frac{121}{4}=\frac{245}{4} }
Radius of C is \mathrm{\sqrt{\frac{245}{4}}=\frac{7 \sqrt{5}}{2} } units
and diameter of \mathrm{C=7 \sqrt{5} } units

Posted by

Ajit Kumar Dubey

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