Find the equation of circle if centre -a - -b and radius -a2 - b2-

Centre = ( a, b) radius = √(a^2 b^2)=r ∴ equation of circle = (x ( a))^2+(y ( b))^2=r^2 #160; (x+a)^2+(y+b)^2=a^2 b^2 #160; x ...

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Standard VIII

Mathematics

Equation of circle with (h,k) as center

Find the equa...

Question

Find the equation of circle if centre $(- a, - b)$ and radius $\sqrt{a^{2} - b^{2}}$ .

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Solution

centre = $(- a, - b)$ radius = $\sqrt{a^{2} - b^{2}} = r$
$∴$ equation of circle =
$(x - (- a))^{2} + (y - (- b))^{2} = r^{2}$
$(x + a)^{2} + (y + b)^{2} = a^{2} - b^{2}$
$x^{2} + 2 a x + a^{2} + y^{2} + 2 b y + b^{2} = a^{2} - b^{2}$
$x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0$
$∴$ equation of circle is $x^{2} + y^{2} + 2 a x + 2 b y - 2 b^{2} = 0$

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Find the equation of circle if centre (−a,−b) and radius √a2−b2.

centre = (−a,−b) radius = √a2−b2=r∴ equation of circle =(x−(−a))2+(y−(−b))2=r2 (x+a)2+(y+b)2=a2−b2 x2+2ax+a2+y2+2by+b2=a2−b2 x2+y2+2ax+2by+2b2=0∴ equation of circle is x2+y2+2ax+2by−2b2=0

Find the equation of circle if centre $(- a, - b)$ and radius $\sqrt{a^{2} - b^{2}}$ .