Question

In the following find the equation of the circle with centre (-a, -b) and radius $\sqrt{a^{2}-b^{2}}$.

Answer 1

here is answer take the following step( s) into consideration !!

r = √a² - b²

√a²- b²= √ a²+ b² - c

c = 2b ²

so , general equation of circle give by

x ²+ y² + 2gx + 2fy +2b²

so we have point => ( -a ,- b )

x² + y² -2g -2f + 2b² = 0

hope it help you !!

thanks!!

Answer 2

Answer:

$x^{2}+y^{2} +2ax +2by+2b^{2} =0$

Step-by-step explanation:

To find the equation of the circle with center (-a,-b) and radius $\sqrt{a^{2}-b^{2}}$

$(x-m)^{2} +(y-n)^{2} =r^{2}$ centre at (m,n) and radius is r.

Now put the given values of centre and radius in standard equation

$\\ (x+a)^{2} +(y+b)^{2} =(\sqrt{a^{2}-b^{2}})^{2}\\ \\ \\ x^{2}+a^{2} +2ax+y^{2} +b^{2}+2by=a^{2}-b^{2}\\ \\ \\ x^{2}+y^{2}+2ax +2by+2b^{2} =0$

is the euation of the circle having centre at (-a,-b) and radius is $\sqrt{a^{2}-b^{2}}$.