Match each circle construction to its point. Column I: (A) Point where x + 2 y = 5 touches x 2 + y 2 − 4 x − 8 y + 15 = 0 . (B) O = ( 0 , 0 ) ; O A , OB are tangents to x 2 + y 2 − 4 x − 8 y + 1 = 0 .

Question

Match each circle construction to its point. Column I: (A) Point where $x + 2 y = 5$ touches $x^{2} + y^{2} - 4 x - 8 y + 15 = 0$ . (B) $O = (0, 0); O A, OB$ are tangents to $x^{2} + y^{2} - 4 x - 8 y + 1 = 0$ . Circumcentre of $△ O A B$ . (C) Intersection of transverse common tangents of $x^{2} + y^{2} - 2 x + 4 y + 4 = 0$ and $x^{2} + y^{2} + 4 x - 2 y + 1 = 0$ . (D) Centre of a circle through $(0, 0), (1, 0)$ and tangent to $x^{2} + y^{2} = 9$ . Column II: (p) $(1/2, 2) (q) (1, 2) (r) (1/2, - 2) (s) (0, - 1)$

Column I	Column II
(A) Tangency point of x+2y=5 and x²+y²-4x-8y+15=0	(p) (1/2, √2)
(B) Circumcentre of △OAB, OA and OB tangent to x²+y²-4x-8y+1=0	(q) (1,2)
(C) Transverse common tangents intersection for x²+y²-2x+4y+4=0 and x²+y²+4x-2y+1=0	(r) (1/2, -√2)
(D) Centre of circle through (0,0),(1,0), tangent to x²+y²=9	(s) (0,-1)

TestHub · Accepted Answer

A: The circle has centre (2,4) and radius sqrt(5); projecting onto x+2y=5 gives (1,2). B: With the corrected +1, the circle has centre (2,4) and radius sqrt(19). Since O is outside, OA and OB are real; their tangent points lie on the circle with diameter OC. Thus the circumcentre of triangle OAB is midpoint (1,2), matching q. C: The internal similitude centre of circles centred at (1,-2),(-2,1) with radii 1,2 is (0,-1), matching s. D: The two centres (1/2, sqrt(2)), (1/2,-sqrt(2)) follow from the stated tangency and passing through (0,0),(1,0), matching p,r.

Mathematics - Circle Question with Solution | TestHub

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