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The tangent to the circle x^2+y^2=5 at (...

The tangent to the circle `x^2+y^2=5` at `(1,-2)` also touches the circle `x^2+y^2-8x+6y+20=0` . Find the coordinats of the corresponding point of contact.

Text Solution

Verified by Experts

Equation of tangent to `x^(2)+y^(2)=5` at (1,-2) is `x-2y-5=0`.
Solving this with the second circle, we get
`(2y+5)^(2)+y^(2)-8(2y+5)+6y+20=0`
`implies 5y^(2)+10y+5=0`
`implies (y+1)^(2)=0`
`implies y=-1`
`implies x =-2+5=3`
Thus, point of contact on second circle is `(3,-1)`.
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