Prove that the tangent to the circle $x^{2} + y^{2} = 5$ at the point $(1, - 2)$ also touches the circle $x^{2} + y^{2} - 8 x + 6 y + 20 = 0$ and find its point of contact.

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Solution

$T_{(1, - 2)} = x - 2 y - 5 = 0$
Apply $p = r$ with $(4, - 3), \sqrt{5}$ . It is satisfied. Point of contact as in part (a) is $(3, - 1)$ .

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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

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

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