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Question
Answer the following :
Find the equation of tangent to Circle x2 + y2 – 6x – 4y = 0, at the point (6, 4) on it
Solution 1
Given equation of the circle is
x2 + y2 – 6x – 4y = 0
Comparing this equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
2g = –6, 2f = –4, c = 0
∴ g = –3, f = –2, c = 0
The equation of a tangent to the circle
x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
∴ the equation of the tangent at (6, 4) is
x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0
∴ 6x + 4y – 3x – 18 – 2y – 8 = 0
∴ 3x + 2y – 26 = 0
Solution 2
Given equation of the circle is
x2 + y2 – 6x – 4y = 0
∴ x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.
Slope of OP = `(4 - 0)/(6 - 0) = 2/3`
Since, OP is perpendicular to the required tangent.
∴ Slope of the required tangent = `(-3)/2`
∴ the equation of the tangent at (6, 4) is
y – 4 = `(-3)/2(x - 6)`
∴ 2(y – 4) = –3(x – 6)
∴ 2y – 8 = –3x + 18
∴ 3x + 2y – 26 = 0
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