Question

The equation of a circle which cuts the three circles

$x^2+y^2-3x-6y+14=0,$

$x^2+y^2-x-4y+8=0$

$x^2+y^2+2x-6y+9=0$

orthogonally is $\underset{–}{}$

• A. x² + y² - 2x - 4y + 1 = 0
• B. x² + y² + 2x + 4y + 1 = 0
• C. x² + y² - 2x + 4y + 1 = 0
• D. x² + y² - 2x - 4y - 1 = 0

Answer 1

The correct option is A

x² + y² - 2x - 4y + 1 = 0

The required circle should have center at the radical center of the three given circles and radius equal to length of the tangent from it to any of three circles cuts all the three circles orthogonally.

Given circles are

$S_1=0,x^2+y^2-3x-6y+14=0$- - - - - - (1)

$S_2=0,x^2+y^2-x-4y+8=0$- - - - - - (2)

$S_3=0,x^2+y^2+2x-6y+9=0$ - - - - - - (3)

Radical axes of circle (1) and (2) is

$S_1-S_2=0$

$2x+2y=6$

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

Radical axes of circle (2) and (3) is

$S_2-S_3=0$

$3x-2y+1=0$- - - - - - (5)

Solving equation (4) and (5)

We get, $x=1,y=2$

Thus, coordinates of radical center is (1, 2)

Since, radius of the required circle = length of the tangent

so, length of the tangent from (1,2) is $\sqrt{s_1}$

$r=\sqrt{(1{)}^2+(2{)}^2-3\times 1-6\times 2+14}$

$\sqrt{1+4-3-12+14}=\sqrt{4}=2$

Hence, equation of required circle is

$(x-1{)}^2+(y-2{)}^2=22$

$x^2+1-2x+y^2+4-4y-4=0$

$x^2+y^2-2x-4y+1=0$

Option A is correct