The equation of the circle which cuts the circles S1 ≡ x2+y2-4=0 S2 ≡ x2+y2-6 x-8 y+10=0 S3 ≡ x2+y2+2 x-4 y-2=0 at the extremities of diameters of these circles is

Question

The equation of the circle which cuts the circles S1 ≡ x2+y2-4=0 S2 ≡ x2+y2-6 x-8 y+10=0 S3 ≡ x2+y2+2 x-4 y-2=0 at the extremities of diameters of these circles is (A) x2+y2-4 x-6 y-4=0 (B) x2+y2+4 x-4=0 (C) x2+y2=25 (D) x2+y2+x+y+1=0

Tardigrade Admin · Accepted Answer

Let the equation of required circle is S ≡ x2+y2+2 g x+2 f y+c=0 and equation of given circles S1 ≡ x2+y2-4=0 S2 ≡ x2+y2-6 x-8 y+10=0 S3=x2+y2+2 x-4, y-2=0 ∵ Circle S=0 cuts the circles S1=0, S2=0 and S3=0 at the extermities of the diameters, so common chord of S=0 and S1=0 passes through the centre of the circle S1=0, so c=-4 Similarly (2 g+6) x+(2 f+8) y-14=0 passes through (3,4), so 6 g+8 f+36=0 ⇒ 3 g+4 f+18=0 ldots(i) and (2 g-2) x+(2 f+4) y-2=0, passes through (-1,2), so -2 g+4 f+8=0 ..........(ii) From Eqs. (i) and (ii), we get (g, f)=(-2,-3) So equation of required circle is x2+y2-4 x-6 y-4=0

Q. The equation of the circle which cuts the circles
$S_{1} \equiv x^{2} + y^{2} - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} - 6 x - 8 y + 10 = 0$
$S_{3} \equiv x^{2} + y^{2} + 2 x - 4 y - 2 = 0$
at the extremities of diameters of these circles is

$x^{2} + y^{2} - 4 x - 6 y - 4 = 0$

$x^{2} + y^{2} + 4 x - 4 = 0$

$x^{2} + y^{2} = 25$

$x^{2} + y^{2} + x + y + 1 = 0$

Q. The equation of the circle which cuts the circles S1​≡x2+y2−4=0 S2​≡x2+y2−6x−8y+10=0 S3​≡x2+y2+2x−4y−2=0 at the extremities of diameters of these circles is

x2+y2−4x−6y−4=0

x2+y2+4x−4=0

x2+y2=25

x2+y2+x+y+1=0

Q. The equation of the circle which cuts the circles
$S_{1} \equiv x^{2} + y^{2} - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} - 6 x - 8 y + 10 = 0$
$S_{3} \equiv x^{2} + y^{2} + 2 x - 4 y - 2 = 0$
at the extremities of diameters of these circles is

$x^{2} + y^{2} - 4 x - 6 y - 4 = 0$

$x^{2} + y^{2} + 4 x - 4 = 0$

$x^{2} + y^{2} = 25$

$x^{2} + y^{2} + x + y + 1 = 0$