Tangent to circle x^2 + y^2 = 5 at point (1, -2) also touches circle x^2 + y^2 -8x + 6y + 20 = 0 . T

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10-1.Circle and System of Circles

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Tangent to the circle $x^2 + y^2$ = $5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 -8x + 6y + 20$ = $0$ . Then its point of contact is

A

$(-2,1)$

B

$(-1,-1)$

C

$(-3,0)$

D

$(3,-1)$

Solution

Tangent at $(1,-2)$ to ${x^2} + {y^2} = 5$ is

$x-2y-5=0$ …….$(1)$

This tangent touchescircle

${x^2} + {y^2} – 8x + 6y + 20 = 0$ at $\left( {{x_1},{y_1}} \right)$

which is food of perpendicular from center $(4,-3)$ upon tangent i.e. $(3,-1)$

Standard 11

Mathematics

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