The equation of the normal at (1, 1) to the c | vedclass

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(A) The equation of the circle is $x^2 + y^2 - x - 3y - 4 = 0$. To find the slope of the tangent at $(1, 1)$,we differentiate the equation with respec

Vedclass · Accepted Answer

$x + y - 2 = 0$

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(A) The equation of the circle is $x^2 + y^2 - x - 3y - 4 = 0$. To find the slope of the tangent at $(1, 1)$,we differentiate the equation with respect to $x$: $2x + 2yy' - 1 - 3y' = 0$ $y'(2y - 3) = 1 - 2x$ $y' = \frac{1 - 2x}{2y - 3}$ At the point $(1, 1)$,the slope of the tangent $m_T$ is: $m_T = \frac{1 - 2(1)}{2(1) - 3} = \frac{-1}{-1} = 1$. The slope of the normal $m_N$ is given by $m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1$. The equation of the normal at $(1, 1)$ is: $y - 1 = -1(x - 1)$ $y - 1 = -x + 1$ $x + y - 2 = 0$.

The equation of the normal at $(1, 1)$ to the circle $x^2 + y^2 - x - 3y - 4 = 0$ is

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