Maths Solve the Following Question.(3 Marks)

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

$a^2=16$

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

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

Here, $m=-2, a^2=16$

the required equations of the tangents are

$\begin{aligned} & y=-2 x \pm \sqrt{16\left[1+(-2)^2\right]} \\ & \Rightarrow y=-2 x \pm \sqrt{16(5)} \\ & \Rightarrow y=-2 x \pm 4 \sqrt{ } 5 \\ & \Rightarrow 2 x+y \pm 4 \sqrt{ } 5=0\end{aligned}$

$x^2+y^2=r^2$ at $\left(x_1, y_1\right)$ is $x x_1+y y_1=r^2$

Equation of the tangent at (3, -4) is

x(3) + y(-4) = 25

⇒ 3x – 4y – 25 = 0 ……(i)

Given equation of circle is $x^2+y^2+8 x-4 y+c=0$

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

2g = 8, 2f = -4

⇒ g = 4, f = -2

$\therefore C=(-4,2)$ and $r=\sqrt{4^2+(-2)^2-c}=\sqrt{20-c}$

Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

$\begin{aligned} & \left|\frac{3(-4)+(-4)(2)-25}{\sqrt{3^2+4^2}}\right|=\sqrt{20-c} \\ & \left|\frac{-45}{\sqrt{25}}\right|=\sqrt{20-c} \\ & \left|\frac{-45}{5}\right|=\sqrt{20-c}\end{aligned}$

$\begin{aligned} & |-9|=\sqrt{20-c} \\ & 81=20-c \\ & c=-61\end{aligned}$

...[Squaring both the sides]

$x^2+y^2-4 x+6 y=1$ i.e, $x^2+y^2-4 x+6 y-1=0$

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

2g = -4, 2f = 6

⇒ g = -2, f = 3

Centre of the circle = (-g, -f) = (2, -3)

Given circle is concentric with the required circle.

∴ They have same centre.

∴ Centre of the required circle = (2, -3)

The equation of a circle with centre at $(h, k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$

Here, h = 2, k = -3 and r = 4

∴ the required equation of the circle is

$\begin{aligned} & (x-2)^2+[y-(-3)]^2=4^2 \\ & \Rightarrow(x-2)^2+(y+3)^2=16 \\ & \Rightarrow x^2-4 x+4+y^2+6 y+9-16=0 \\ & \Rightarrow x^2+y^2-4 x+6 y-3=0\end{aligned}$

x = 5 cos θ and y = 5 sin θ

$\begin{aligned} & \Rightarrow x^2+y^2=25 \cos ^2 \theta+25 \sin ^2 \theta \\ & \Rightarrow x^2+y^2=25\left(\cos ^2 \theta+\sin ^2 \theta\right) \\ & \Rightarrow x^2+y^2=25(1)=25\end{aligned}$

The equation of the director circle of the circle $x^2+y^2=a^2$ is $x^2+y^2=2 a^2$.

Here, a = 5

∴ the required equation is

$\begin{aligned} & x^2+y^2=2(5)^2=2(25) \\ & \therefore x^2+y^2=50\end{aligned}$

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

2g = -4, 2f = -2, c = -4

⇒ g = -2, f = -1, c = -4

The equation of a tangent to the circle

$x^2+y^2+2 g x+2 f y+c=0$ at $\left(x_1, y_1\right)$ is $x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$

the equation of the tangent at (-1, 1) is

⇒ x(-1) + y(1) – 2(x – 1) – 1(y + 1) – 4 = 0

⇒ -3x – 3 = 0

⇒ -x – 1 = 0

⇒ x = -1

∴ x = -1 is the tangent to the given circle at (-1, 1).

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on

the Y-axis at point B.

∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).

Since ∠BOA is a right angle.

AB represents the diameter of the circle.

The equation of a circle having $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ as endpoints of diameter is given by

$\begin{aligned} & \left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0 \\ & \text { Here, } x_1=4_2 y_1=0, x_2=0, y_2=6\end{aligned}$

∴ the required equation of the circle is ⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0

$\begin{aligned} & \Rightarrow x^2-4 x+y^2-6 y=0 \\ & \Rightarrow x^2+y^2-4 x-6 y=0\end{aligned}$