Point of contact of tangent to circle {x^2} + {y^2} = 5 at point (1, -2) which touches circle {x^2}

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10-1.Circle and System of Circles

hard

The point of contact of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1, -2)$ which touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$, is

A

$(2, -1)$

B

$(3, -1)$

C

$(4, -1)$

D

$(5, -1)$

Solution

(b) Equation of the tangent at $(1,\; – 2)$ to the circle ${x^2} + {y^2} = 1$ is $x – 2y = 5$.

Here, only point $(3, -1)$ lies on the tangent.

Standard 11

Mathematics

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