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VITEEE 2010 Paper

Question : 119 of 120

Marks: +1, -0

If two circles

2x^{2}+2y^{2}-3x+6y+k=0

x^{2}+y^{2}-4x+10y+16=0

cut orthogonally, then the value of k is

Solution:

The given two circles are

2x^{2}+2y^{2}-3x+6y+k=0

⇒

x^{2}+y^{2}-\frac{3}{2}x+3y+\frac{k}{2}=0

...(i)

and

x^{2}+y^{2}-4x+10y+16=0

...(ii)

Since, general equation of circle is

x^{2}+y^{2}+2gx+2fy+c=0

...(iii)

Therefore, comparing eqs. (i) and (ii) with eq. (iii), we get

g_{1}=-\frac{3}{4},f_{1}=\frac{3}{2},c_{1}=\frac{k}{2}

and

g_{2}=-2,f_{2}=5,c_{2}=16

Both the circles cut orthogonally,

∴

2(g_{1}g_{2}+f_{1}f_{2})=c_{1}+c_{2}

⇒

2\left(\frac{3}{2}+\frac{15}{2}\right)=\frac{k}{2}+16

⇒

18=\frac{k}{2}+16

⇒

\frac{k}{2}=2

⇒

k=4

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