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Question
Answer the following :
Find the equations of the tangents to the circle x2 + y2 – 2x + 8y – 23 = 0 having slope 3
Solution
Let the equation of the tangent with slope 3 be y = 3x + c
i.e., 3x – y + c = 0 ...(1)
The centre of the circle x2 + y2 – 2x + 8y – 23 = 0 is
C (1, – 4) and its radius = `sqrt(1 + 16 + 23) = sqrt(40)`.
Since line (1) is tangent to the circle, the length of perpendicular from the centre C (1, - 4) to the line is equal to radius.
∴ `|(3(1) + (-1)(-4) + "c")/sqrt(3^2 + (-1)^2)| = sqrt(40)`
∴ `|(3 + 4 + "c")/sqrt(10)| = sqrt(40)`
∴ 7 + c = ± 20
∴ 7 + c = 20 or 7 + c = – 20
∴ c = 13 or c = – 27
∴ from (1), the equations of the tangents are
3x – y + 13 = 0 and 3x – y – 27 = 0.
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