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Equation of Tangent in Point Form

The point of ...

Question

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

(2,-1)

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(3,-1)

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(4,-1)

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(5,-1)

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Solution

The correct option is B (3,-1)
Equation of the tangent at (1,-2) to the circle $x^{2} + y^{2} = 1$ is $x - 2 y = 5$ .
Here, only point (3,-1) lies on the tangent.

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Similar questions

x^{2} + y^{2} = 5

(1, - 2)

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

x^{2} + y^{2} = 5

(1, - 2)

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

x^{2} + y^{2} = 5

(1, - 2)

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

x^{2} + y^{2} = 5

(1, - 2)

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

x^{2} + y^{2} = 5

(1, - 2)

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

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