Q.9. To find equation of a circle through origin and contact points of tangents from origin to a given circle

Question 9:

Tangents OA and OB are drawn from the origin to the circle (x-2)^2+(y-2)^2=1 then prove that the equation of the circumcircle of the triangle OAB is x^2+y^2-2x-2y=0.

Answer 9:

The given circle has centre (2,2) and radius = 1. Two tangents can be drawn from origin touching the circle at A and B.

Let C = centre of the circle.

Now quadrilateral OACB is concyclic because ∠OAC + ∠OBC = 90 + 90 = 180 degrees.

Therefore, the required circle passing through O, A and B also passes through C.

Also, as ∠OAC = 90 degrees, OC is the diameter of the circle through O, A and B.

Thus, extremities of the diameter of the required circle are (0, 0) and (2, 2)

=> equation of the required circle is

(x-0)(x-2) + (y-0)(y-2) = 0

=> x^2 + y^2 - 2x - 2y = 0.

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