Maths Solve the Following Question.(4 Marks)

3x – y + c = 0 ……(i)

Given equation of circle is $x^2+y^2-2 x+8 y-23=0$

Comparing this equation with $x^2+y^2+2 g x+2 f y+c=0$, we get

2g = -2, 2f = 8, c = -23

g = -1, f = 4, c = -23

The centre of the circle is C(1, -4)

and its radius $=\sqrt{1+16+23}$

= √40

= 2√10

Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.

$\begin{aligned} & \left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|=2 \sqrt{ } 10 \\ & \Rightarrow\left|\frac{7+c}{\sqrt{10}}\right|=2 \sqrt{ } 10 \\ & \Rightarrow(7+c)= \pm 20 \\ & \Rightarrow 7+c=20 \text { or } 7+c=-20 \\ & \Rightarrow c=13 \text { or } c=-27\end{aligned}$

∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Comparing this equaiton with $x^2+y^2=a^2$, we get

$a^2=36$

Given equation of line is 5x + y = 2

Slope of this line = -5

Since,the required tangents are perpendicular to the given line.

Slope of required tangents $(\mathrm{m})=\frac{1}{5}$

Equations of the tangents to the circle $x^2+y^2=a^2$ with slope $m$ are

$y=m x \pm \sqrt{a^2\left(1+m^2\right)}$

the required equations of the tangents are

$\begin{aligned} y & =\frac{1}{5} x \pm \sqrt{36\left[1+\left(\frac{1}{5}\right)^2\right]} \\ & =\frac{1}{5} x \pm \sqrt{36\left(1+\frac{1}{25}\right)}\end{aligned}$

$\begin{aligned} & y=\frac{1}{5} x \pm \frac{6}{5} \sqrt{26} \\ & 5 y=x \pm 6 \sqrt{26} \\ & x-5 y \pm 6 \sqrt{26}=0\end{aligned}$

⇒ y = 2x + 6

Substituting $y=2 x+6$ in $x^2+y^2+10 x+9=0$, we get

$\begin{aligned} & \Rightarrow x^2+(2 x+6)^2+10 x+9=0 \\ & \Rightarrow x^2+4 x^2+24 x+36+10 x+9=0 \\ & \Rightarrow 5 x^2+34 x+45=0 \\ & \Rightarrow 5 x^2+25 x+9 x+45=0 \\ & \Rightarrow(5 x+9)(x+5)=0 \\ & \Rightarrow 5 x=-9 \text { or } x=-5 \\ & \Rightarrow x=\frac{-9}{5} \text { or } x=-5\end{aligned}$

When $x=\frac{-9}{-9}$

$\begin{aligned} & y=2 \times \frac{-9}{5}+6 \\ & =\frac{-18}{5}+6 \\ & =\frac{-18+30}{5} \\ & =\frac{12}{5}\end{aligned}$

$\therefore$ Point of intersection is $A\left(\frac{-9}{5}, \frac{12}{5}\right)$

When x = -5, y = -10 + 6 = -4

∴ Point of intersection in B (-5, -4).

By diameter form, equation of circle with AB as diameter is

$\begin{aligned} & \left(x+\frac{9}{5}\right)(x+5)+\left(y-\frac{12}{5}\right)(y+4)=0 \\ & \Rightarrow(5 x+9)(x+5)+(5 y-12)(y+4)=0 \\ & \Rightarrow 5 x^2+25 x+9 x+45+5 y^2+20 y-12 y-48=0 \\ & \Rightarrow 5 x^2+5 y^2+34 x+8 y-3=0\end{aligned}$

3x – y + c = 0 ……(i)

Given equation of circle is $x^2+y^2-2 x+8 y-23=0$

Comparing this equation with $x^2+y^2+2 g x+2 f y+c=0$, we get

2g = -2, 2f = 8, c = -23

g = -1, f = 4, c = -23

The centre of the circle is C(1, -4)

and its radius $=\sqrt{1+16+23}$

= √40

= 2√10

Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.

$\begin{aligned} & \left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|=2 \sqrt{ } 10 \\ & \Rightarrow\left|\frac{7+c}{\sqrt{10}}\right|=2 \sqrt{ } 10 \\ & \Rightarrow(7+c)= \pm 20 \\ & \Rightarrow 7+c=20 \text { or } 7+c=-20 \\ & \Rightarrow c=13 \text { or } c=-27\end{aligned}$

∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Comparing this equaiton with $x^2+y^2=a^2$, we get

$a^2=36$

Given equation of line is 5x + y = 2

Slope of this line = -5

Since,the required tangents are perpendicular to the given line.

Slope of required tangents $(\mathrm{m})=\frac{1}{5}$

Equations of the tangents to the circle $x^2+y^2=a^2$ with slope $m$ are

$y=m x \pm \sqrt{a^2\left(1+m^2\right)}$

the required equations of the tangents are

$\begin{aligned} y & =\frac{1}{5} x \pm \sqrt{36\left[1+\left(\frac{1}{5}\right)^2\right]} \\ & =\frac{1}{5} x \pm \sqrt{36\left(1+\frac{1}{25}\right)}\end{aligned}$

$\begin{aligned} & y=\frac{1}{5} x \pm \frac{6}{5} \sqrt{26} \\ & 5 y=x \pm 6 \sqrt{26} \\ & x-5 y \pm 6 \sqrt{26}=0\end{aligned}$

⇒ y = 2x + 6

Substituting $y=2 x+6$ in $x^2+y^2+10 x+9=0$, we get

$\begin{aligned} & \Rightarrow x^2+(2 x+6)^2+10 x+9=0 \\ & \Rightarrow x^2+4 x^2+24 x+36+10 x+9=0 \\ & \Rightarrow 5 x^2+34 x+45=0 \\ & \Rightarrow 5 x^2+25 x+9 x+45=0 \\ & \Rightarrow(5 x+9)(x+5)=0 \\ & \Rightarrow 5 x=-9 \text { or } x=-5 \\ & \Rightarrow x=\frac{-9}{5} \text { or } x=-5\end{aligned}$

When $x=\frac{-9}{-9}$

$\begin{aligned} & y=2 \times \frac{-9}{5}+6 \\ & =\frac{-18}{5}+6 \\ & =\frac{-18+30}{5} \\ & =\frac{12}{5}\end{aligned}$

$\therefore$ Point of intersection is $A\left(\frac{-9}{5}, \frac{12}{5}\right)$

When x = -5, y = -10 + 6 = -4

∴ Point of intersection in B (-5, -4).

By diameter form, equation of circle with AB as diameter is

$\begin{aligned} & \left(x+\frac{9}{5}\right)(x+5)+\left(y-\frac{12}{5}\right)(y+4)=0 \\ & \Rightarrow(5 x+9)(x+5)+(5 y-12)(y+4)=0 \\ & \Rightarrow 5 x^2+25 x+9 x+45+5 y^2+20 y-12 y-48=0 \\ & \Rightarrow 5 x^2+5 y^2+34 x+8 y-3=0\end{aligned}$