A tangent to circle {x^2} + {y^2} = 5 at point (1,-2) circle {x^2} + {y^2}

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10-1.Circle and System of Circles

medium

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$

Touches

Cuts at real points

Cuts at imaginary points

None of these

(IIT-1975)

(a) Tangent is $x – 2y – 5 = 0$ and points of intersection with circle ${x^2} + {y^2} – 8x + 6y + 20 = 0$ are given by

$4{y^2} + 25 + 20y + {y^2} – 16y – 40 + 6y – 20 = 0$

$ \Rightarrow 5{y^2} + 10y + 5 = 0$

$ \Rightarrow y = – 1$ and $x = – 3$

Standard 11

Mathematics

normal

hard

normal

normal

hard

(IIT-1985)