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Find the equation of the tangent to the circle x2 + y2 − 2x − 4y + 3 = 0 at the point (2, 3). |
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Answer» We have 22 + 32 - 2(2) - 4(3) + = 16 - 16 = 0 which implies that (2, 3) lies on the circle S = x2 + y2 − 2x − 4y +3 = 0. Here, (x1, y1) = (2, 3) so that the equation of the tangent at (2, 3) is S1 = x(2) + y(3) − (x + 2) − 2(y + 3) + 3 = 0 that is, S1 = x + y − 5 = 0. |
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