QUESTION

# The equation of the normal to the circle ${x^2} + {y^2} - 8x - 2y + 12 = 0$ at the points whose ordinate is $- 1$, will beA. $2x - y - 7 = 0,2x + y - 9 = 0$B. $2x + y - 7 = 0,2x + y + 9 = 0$C. $2x + y + 7 = 0,2x + y + 9 = 0$D. $2x - y + 7 = 0,2x - y + 9 = 0$

Hint: First of all, find the points at which Normals are formed by substituting the ordinate in the given circle equation. Then find the slopes of the Normals at obtained points. And then find the equations of the normals by using slope point form method. So, use this concept to reach the solution of the given problem.

Complete step-by-step solution -

Given equation of the circle ${x^2} + {y^2} - 8x - 2y + 12 = 0................................\left( 1 \right)$
We have to draw normal whose ordinates is $- 1$ i.e., $y = - 1$
Putting $y = - 1$, in equation (1) we have
$\Rightarrow {x^2} + {\left( { - 1} \right)^2} - 8x - 2\left( { - 1} \right) + 12 = 0 \\ \Rightarrow {x^2} + 1 - 8x + 2 + 12 = 0 \\ \Rightarrow {x^2} - 8x + 15 = 0 \\$
Splitting the terms, we get
$\Rightarrow {x^2} - 5x - 3x + 15 = 0 \\ \Rightarrow x\left( {x - 5} \right) - 3\left( {x - 5} \right) = 0 \\ \Rightarrow \left( {x - 5} \right)\left( {x - 3} \right) = 0 \\ \therefore x = 3,5 \\$
So, the points where Normal to be drawn is $\left( {3, - 1} \right){\text{ and }}\left( {5, - 1} \right)$
Differentiating equation (1) with respective to ‘$x$’
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^2} + {y^2} - 8x - 2y + 12} \right) = \dfrac{d}{{dx}}\left( 0 \right) \\ \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} - 8 - 2\dfrac{{dy}}{{dx}} = 0 \\ \Rightarrow 2x - 8 = 2\dfrac{{dy}}{{dx}} - 2y\dfrac{{dy}}{{dx}} \\ \Rightarrow 2\left( {x - 4} \right) = 2\dfrac{{dy}}{{dx}}\left( {1 - y} \right) \\ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - y} \right) = \left( {x - 4} \right) \\ \therefore \dfrac{{dy}}{{dx}} = \dfrac{{x - 4}}{{1 - y}} \\$
The slope of the Normal at point $\left( {3, - 1} \right)$ is given by
$\Rightarrow \dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{\left( {3, - 1} \right)}}}} = \dfrac{{ - 1}}{{\dfrac{{3 - 4}}{{1 - \left( { - 1} \right)}}}} = \dfrac{{ - 1}}{{\dfrac{{ - 1}}{2}}} = 2$
We know that the equation of the line at point $\left( {{x_1},{y_1}} \right)$with slope $m$is given by $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
So, the normal at point $\left( {3, - 1} \right)$ to the given circle is
$\Rightarrow \left( {y - \left( { - 1} \right)} \right) = 2\left( {x - 3} \right) \\ \Rightarrow y + 1 = 2x - 6 \\ \therefore 2x - y - 7 = 0 \\$
The slope of the Normal at point $\left( {5, - 1} \right)$is given by
$\Rightarrow \dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{\left( {5, - 1} \right)}}}} = \dfrac{{ - 1}}{{\dfrac{{5 - 4}}{{1 - \left( { - 1} \right)}}}} = \dfrac{{ - 1}}{{\dfrac{1}{2}}} = - 2$
So, the normal at point $\left( {5, - 1} \right)$ is given by
$\Rightarrow \left( {y - \left( { - 1} \right)} \right) = - 2\left( {x - 5} \right) \\ \Rightarrow y + 1 = - 2x + 10 \\ \therefore 2x + y - 9 = 0 \\$
Thus, the correct option is A. $2x - y - 7 = 0,2x + y - 9 = 0$

Note: The ordinate is the y-coordinate of a point on the coordinate plane. The x-coordinate of a point is called the abscissa. If the slope of the tangent at point $\left( {{x_1},{y_1}} \right)$ is given by ${\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}}$ and the slope of the normal is given by $\dfrac{{ - 1}}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{\left( {{x_1},{y_1}} \right)}}}}$.