Question

A circle cutting the circle x2+y2 =4 orthogonally and having its centre on the line 2x - 2y + 9 = 0, passesoh two fixed points. These points are(1) (4,0) and (0,4)(2) (-4, 4) and (-1/2, 1/2) (3) (4, -4) and (1/2, -1/2) (4) (-4,0) and (4,0)​

Answer 1

These points are,

$\left ( \dfrac{-1}{2},\dfrac{1}{2} \right ) \ And \ \left ( -4, 4 \right )$

Step-by-step explanation:

Assume a circle 'C' cutting the circle $x^2 + y^2 = 4$orthogonally.

Let the equation of the circle C is:

• $x^2 + y^2 + 2gx + 2fy + c = 0$

• We know that its center is at (-g, -f).

We can understand that the center (-g, -f) would lie on 2x - 2y +9 = 0

• So, 2g - 2f + 9 = 0

2g = 2f + 9

• Since, it cuts the circle $x^2 + y^2 = 4$ orthogonally, we can write as:

$(2g\times 0 ) + (2f\times 0 ) = c - 4$

By solving it, we get,

c = 4

Now, we write as;

• $x^2 + y^2 + (2f + 9)x + 2fy + 4 = 0$

$(x^2 + y^2 + 9x +9) + f(2x + 2y) = 0$

• line 2x + 2y = 0 cuts the circle $x^2 + y^2 + 9x + 4 = 0$

So, by solving both equations, we can find out their intersection points.

Thus, by solving them, we get two points as;

$\left ( \dfrac{-1}{2},\dfrac{1}{2} \right ) \ And \ \left ( -4, 4 \right )$