Equation of circle with centre (-a, - b) and radius √ a2-b2 is

Question

Equation of circle with centre (-a, - b) and radius √ a2-b2 is (A) x2 + y2 - 2ax- 2by- 2b2 = 0 (B) x2 + y2 - 2ax+ 2by- 2a2 = 0 (C) x2 + y2 + 2ax+ 2by+2b2 = 0 (D) x2 + y2 - 2ax- 2by+ 2b2 = 0

Tardigrade Admin · Accepted Answer

We know that equation of circle having centre

(h, k)

and radius

r

is

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

.
Here, centre

= (- a, - b)

and radius

= a^{2} - b^{2}

∴

Required equation of circle is

(x + a)^{2} + (y + b)^{2} = (a^{2} - b^{2})^{2}

\Rightarrow x^{2} + a^{2} + 2 a x + y^{2} + b^{2} + 2 b y = a^{2} - b^{2}

\Rightarrow x^{2} + 2 a x + y^{2} + 2 b y + 2 b^{2} = 0

Q. Equation of circle with centre $(- a, - b)$ and radius $a^{2} - b^{2}$ is

$x^{2} + y^{2} - 2 a x - 2 b y - 2 b^{2} = 0$

$x^{2} + y^{2} - 2 a x + 2 b y - 2 a^{2} = 0$

$x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0$

$x^{2} + y^{2} - 2 a x - 2 b y + 2 b^{2} = 0$

Q. Equation of circle with centre (−a,−b) and radius a2−b2​ is

x2+y2−2ax−2by−2b2=0

x2+y2−2ax+2by−2a2=0

x2+y2+2ax+2by+2b2=0

x2+y2−2ax−2by+2b2=0

We know that equation of circle having centre (h,k) and radius r is (x−h)2+(y−k)2=r2. Here, centre =(−a,−b) and radius =a2−b2​ ∴ Required equation of circle is (x+a)2+(y+b)2=(a2−b2​)2 ⇒x2+a2+2ax+y2+b2+2by=a2−b2 ⇒x2+2ax+y2+2by+2b2=0

Q. Equation of circle with centre $(- a, - b)$ and radius $a^{2} - b^{2}$ is

$x^{2} + y^{2} - 2 a x - 2 b y - 2 b^{2} = 0$

$x^{2} + y^{2} - 2 a x + 2 b y - 2 a^{2} = 0$

$x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0$

$x^{2} + y^{2} - 2 a x - 2 b y + 2 b^{2} = 0$