Question

The equation of the circle with centre (3, 2)
and the power of (1, -2) w.r.t the circle
x2 + y2 = 1 as radius is
1) x² + y² - 6x – 4y – 3=0
2) x2 + y2 – 3x – 2y - 3 = 0
3) x2 + y2 +6x + 4y – 3 = 0
4 x² + y² - 6x – 4y+3=0​

Answer 1

Given:

Radius of circle is power of (1, -2) w.r.to circle $x^{2} +y^{2} =1$

Center of the circle is (3, 2)

To find:

Equation of the circle = ?

Solution:

Power of (1, -2) w.r.to circle $x^{2} +y^{2} =1$ can be calculated by putting the point in the circle.

i.e.

$p = x^2+y^2-1\\\Rightarrow p = 1^2 +(-2)^2-1 = 4$

Given that this power is equal to the radius of required circle.

So, radius of circle = 4

Equation of a circle with radius $r$ and center $(h,k)$ can be given as:

$(x-h)^2+(y-k)^2=r^2$

Putting the values:

$(x-3)^2+(y-2)^2=4^2\\\Rightarrow x^2+9-6x+y^2-4y+4=16\\\Rightarrow x^2 + y^2 - 6x- 4y - 3=0$

Therefore the answer is:

1) $x^2 + y^2 - 6x- 4y - 3=0$