If the line px+qy =1 touches the circle x^2+y^2=r^2, prove that the point (p,q) lies on the circle x^2+y^2=r^-2.

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Please find below the solution to the asked query:

We have:px+qy-1=0Equation of circle is x2+y2=r2Center is 0,0 aand radius is r units.As line touches circle, hence length of perpendicular from centre of cirlceto the line must be equal to the radius of circle.p×0+q×0-1p2+q2=r-1p2+q2=r1p2+q2=rSquarring both sides we get:1p2+q2=r2p2+q2=1r2p2+q2=r-2If we replace p with x and q with y, then:x2+y2=r-2Hence point p,q lies on cirlce x2+y2=r-2
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