Question

The equation of the normal to the circle ${{x}^{2}}+{{y}^{2}}-4x+4y-17=0$ which passes through $\left( 1,1 \right)$ is ……

Answer 1

Hint: Any normal of a circle always passes through its center. First find the center of circle by comparing the given equation of circle with the standard equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ for which center of circle is (-g,-f). Then find the equation of the straight line by using two point forms.
Complete step by step solution
We have from question
Equation of circle given ${{x}^{2}}+{{y}^{2}}-4x+4y-17=0$
On comparing the above equation with the standard equation of circle
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
We get
$\begin{aligned} & g=-2 \\\ & f=2 \\\ & c=-17 \\\ \end{aligned}$
Now center of circle

& {{C}_{1}}=(-g,-f) \\\ & \Rightarrow {{C}_{1}}=(-(-2),-(2)) \\\ & \Rightarrow {{C}_{1}}=(2,-2) \\\ \end{aligned}$$ Now we have to find out equation of straight line which passes through $\left( 2,-2 \right)$ and$\left( 1,1 \right)$ Equation of line joining the two points$\left( {{x}_{1,}}{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ can be written as $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ So equation of line joining the point$\left( 2,-2 \right)$ and $\left( 1,1 \right)$ $\begin{aligned} & \Rightarrow y-(-2)=\frac{1-(-2)}{1-2}\left( x-2 \right) \\\ & \Rightarrow y+2=\frac{3}{-1}\left( x-2 \right) \\\ & \Rightarrow y+2=6-3x \\\ & \Rightarrow y+3x=6-2 \\\ & \Rightarrow 3x+y=4 \\\ \end{aligned}$ Hence equation of normal is $3x+y=4$ Note: By geometry normal are perpendicular to the tangent at the point on the circle and line joining center and point of contact at tangent point is perpendicular to the tangent so normal always passes through the center of the circle. We can find the equation of a straight line by using slope form.