At what point the slope of the tangent to the curve x2 + y2 − 2x − 3 = 0 is zero

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(1, 2), (1, −2) Let (x1, y1) be the required point. \text { Since, the point lie on the curve } . \text { Hence }, {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0 . . . \left( 1 \right) \text { Now }, x^2 + y^2 - 2x - 3 = 0 \Rightarrow 2x + 2y dy/dx - 2 = 0 \therefore dy/dx = 2 - 2x/2y = 1 - x/y \text { Now}, \text{ Slope of the tangent } = \left( dy/dx \right)_\left( x_1 , y_1 \right) =1 - x_1/y_1 \text { Slope of the tangent }=0 ...............(\text {Given }) \therefore 1 - x_1/y_1 = 0 \Rightarrow 1 - x_1 = 0 \Rightarrow x_1 = 1 \text { From (1), we get } {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0 \Rightarrow 1 + {y_1}^2 - 2 - 3 = 0 \Rightarrow {y_1}^2 - 4 = 0 \Rightarrow y_1 = $\pm$ 2 \text { So, the points are }\left( 1, 2 \right)\text { and }\left( 1, - 2 \right).

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