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Question: Find the equation of the tangent at the origin to the circle \({x^2} + {y^2}\) - 4\(x\) - 10\(y\) = ...

Find the equation of the tangent at the origin to the circle x2+y2{x^2} + {y^2} - 4xx - 10yy = 0.

Explanation

Solution

Hint: Differentiate the given equation of circle wrt xx to find the expression of dydx\dfrac{{dy}}{{dx}}, This will be the slope of the tangent to the circle, find its value by substituting the point through which it is passing, then at last use point-slope form of the equation of line to get the answer.

Complete step-by-step answer:

The equation given to us by the question is x2+y2−4x−10y=0{x^2} + {y^2} - 4x - 10y = 0. We will differentiate both the sides with respect to x in order to solve this further. We get-
⇒x2+y2−4x−10y=0  ⇒ddx(x2+y2−4x−10y)=ddx0  ⇒2x+2y.dydx−4−10dydx=0  ⇒dydx(2y−10)=4−2x  ⇒dydx=4−2x2y−10  \Rightarrow {x^2} + {y^2} - 4x - 10y = 0 \\\ \\\ \Rightarrow \dfrac{{d}}{{dx}}\left( {{x^2} + {y^2} - 4x - 10y} \right) = \dfrac{{d}}{{dx}}0 \\\ \\\ \Rightarrow 2x + 2y.\dfrac{{dy}}{{dx}} - 4 - 10\dfrac{{dy}}{{dx}} = 0 \\\ \\\ \Rightarrow \dfrac{{dy}}{{dx}}\left( {2y - 10} \right) = 4 - 2x \\\ \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4 - 2x}}{{2y - 10}} \\\
As we know that the value of x and y is (0,0)(0,0), putting this value in the above equation, we get-
⇒dydx=4−2x2y−10  ⇒dydx=4−10=−25  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4 - 2x}}{{2y - 10}} \\\ \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{ - 10}} = - \dfrac{2}{5} \\\
Now, to find out the equation of the tangent, we will-
We know that the value of x1{x_1} and y1{y_1}is (0,0)(0,0) and the value of dydx\dfrac{{dy}}{{dx}} is −25 - \dfrac{2}{5}. So,
We also know that (y−y1)=m(x−x1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right), where m is the slope i.e. dydx\dfrac{{dy}}{{dx}}.
Putting all the values in the above formula, we will get the equation of the tangent-
⇒(y−y1)=m(x−x1)  ⇒(y−0)=−25(x−0)  ⇒y=−25x  ⇒2x+5y=0  \Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) \\\ \\\ \Rightarrow \left( {y - 0} \right) = - \dfrac{2}{5}\left( {x - 0} \right) \\\ \\\ \Rightarrow y = - \dfrac{2}{5}x \\\ \\\ \Rightarrow 2x + 5y = 0 \\\
Hence, the required equation of the tangent is 2x+5y=02x + 5y = 0.

Note: In such questions where you must find the equation of a tangent, do not forget the formula of finding the equation i.e. (y−y1)=m(x−x1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right). Where, m is the slope which you will get by differentiating the given equation of the circle.