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Question
Show that the line 7x − 3y − 1 = 0 touches the circle x2 + y2 + 5x − 7y + 4 = 0 at point (1, 2)
Solution
Given equation of the circle is
x2 + y2 + 5x − 7y + 4 = 0
Comparing this equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
2g = 5, 2f = – 7, c = 4
∴ g = `5/2, "f" = -7/2, "c" = 4`
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 (1, 2) is
`x(1) + y(2) + 5/2(x + 1) - 7/2(y + 2) + 4` = 0
∴ `x + 2y + 5/2x + 5/2 - 7/2y - 7 + 4` = 0
∴ `7/2x - 3/2y - 1/2` = 0
∴ 7x – 3y – 1 = 0, which is same as the given line
∴ The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).
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