Question

Find the equation to the circle which passes through the origin and has its center on the line $x+y+4=0$ and cuts the circle $x^2+y^2-4x+2y+4=0$ orthogonally.

• A. 3x² + 3y² - 4x + 20y = 0
• B. x² + y² + 4x + 20y = 0
• C. 3x² + 3y² + 20 x + 4y = 0
• D. x² + y² + 20x + 4y = 0

Answer 1

The correct option is A

3x² + 3y² - 4x + 20y = 0

Let required circle be

$x^2+y^2+2gx+2fy+c=0$ - - - - - - (1)

Since, this circle passes through origin

Substituting x = 0 and y = 0 in equation (1)

0 + 0 + 0 + 0 + c = 0

c = 0

Required circle equation is

$x^2+y^2+2gx+2fy=0$ - - - - - - (2)

Circles $x^2+y^2+2gx+2fy=0and{x}^2+y^2-4x+2y+4=0$ cuts orthogonally.

Then $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

2(-g) (2) + 2(-f)(-1) = 0 + 4

-4g + 2f = 0

-2g + f = 0 - - - - - - (3)

Centre (-g, -f) lies on the equation x + y + 4 = 0

-g - f + 4 = 0

g + f = 4 - - - - - - (4)

Solving equation 3 and 4

3f = 10

$f=\frac{10}{3},g=4-\frac{10}{3}=\frac{2}{3}$

Substituting the values of g and f in equation (1)

Required equation of the circle is

$x^2+y^2+2\left(\frac{2}{3}\right)x+2\left(\frac{10}{3}\right)y=0$

$3x^2+3y^2+4x+20y=0$

Option A is correct