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Question
Answer the following :
If the tangent at (3, –4) to the circle x2 + y2 = 25 touches the circle x2 + y2 + 8x – 4y + c = 0, find c
Solution
The equation of a tangent to the circle
x2 + y2 = r2 at (x1, y1) is xx1 + yy1 = r2
Equation of the tangent at (3, – 4) is
x(3) + y(–4) = 25
∴ 3x – 4y – 25 = 0 …(i)
Given equation of circle is
x2 + y2 + 8x – 4y + c = 0
Comparing this equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
2g = 8, 2f = – 4
∴ g = 4, f = – 2
∴ C ≡ (– 4, 2) and r = `sqrt(4^2 + (-2)^2 - "c") = sqrt(20 - "c")`
Since line (i) is a tangent to this circle also, the perpendicular distance from C(– 4, 2) to line (i) is equal to radius r.
∴ `|(3(-4) + (-4)(2) - 25)/sqrt(3^2 + 4^2)| = sqrt(20 - "c")`
∴ `|(-45)/sqrt(25)| = sqrt(20 - "c")`
∴ `|(-45)/5| = sqrt(20 - "c")`
∴ `|–9| = sqrt(20 - "c")`
∴ 81 = 20 – c ...[Squaring both the sides]
∴ c = –61
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