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Question

The equation of a circle which touches the lines x=y at origin and passes through the point 2,1 is x2+y2+px+qy=0 Then the value of p,q are


A

5,5

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B

-5,5

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C

5,-5

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D

None of these

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Solution

The correct option is B

-5,5


Explanation of correct answer :

Finding values of p, q :

Let the equation of the circle passing through origin be x2+y2+2gx+2fy=0

It also passes through 2,1.

4+1+4g+2f=04g+2f=-5(1)

Also, the circle touches the line, x=y.

Perpendicular from centre -g,-f to the tangent = Radius

-f+g12+12=g2+f2

Squaring both sides, we get

f2+g2-2fg=2g2+f2f2+g2-2fg=2g2+2f2f2+g2+2fg=0f+g2=0f+g=0g=-f

From eqn (1),

4-f+2f=-5-2f=-5f=52g=-52g=-f

So, the equation of the circle is x2+y2-5x+5y=0

On comparing with,x2+y2+px+qy=0, we get

Thus, p=-5,q=5.

Hence, the correct answer is Option(B).


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