Find-the-equation-to-the-two-circles-each-of-which-touch-the-three-circle-x-2-y-2-4a-2-x-2-y-2-2ax-0-x-2-y-2-2ax-0-

Question Number 77990 by peter frank last updated on 12/Jan/20

F i n d t h e e q u a t i o n t o t h e

t w o c i r c l e s e a c h o f

w h i c h t o u c h t h e t h r e e c i r c l e

x^{2} + y^{2} = 4 a^{2}

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

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

Answered by john santu last updated on 18/Jan/20

(1) x^{2} + y^{2} = {(2 a)}^{2}, C_{1} (0, 0), r_{1} = 2 a

(2) {(x + a)}^{2} + y^{2} = a^{2}, C_{2} (- a, 0), r_{2} = a

(3) x^{2} + {(y - a)}^{2} = a^{2}, C_{3} (0, a), r_{3} = a

T h e c i r c l e e q u a t i o n t h a t w i l l

l o o k f o r i s c e n t e r e d a t p o i n t

(p, q) w i t h r a d i u s r_{4}

\Rightarrow r_{4} = \sqrt{p^{2} + q^{2}}, r_{4} = \sqrt{{(p + a)}^{2} + q^{2}}, r_{4} = \sqrt{p^{2} + {(q - a)}^{2}}

Commented by john santu last updated on 18/Jan/20

\Rightarrow p^{2} + q^{2} = p^{2} + 2 a p + a^{2} + q^{2}

p = - \frac{a}{2} .

\Rightarrow p^{2} + q^{2} = p^{2} + q^{2} - 2 a q + a^{2}

q = \frac{a}{2} . n o w w e c a l c u l a t e

r_{4} = \sqrt{\frac{1}{4} a^{2} + \frac{1}{4} a^{2}} = \frac{1}{2} a \sqrt{2} .

f i n a l l y ∴ {(x + \frac{a}{2})}^{2} + {(y - \frac{a}{2})}^{2} = \frac{a^{2}}{2}

Commented by mr W last updated on 18/Jan/20

s i r :

b u t (r e d) c i r c l e {(x + \frac{a}{2})}^{2} + {(y - \frac{a}{2})}^{2} = \frac{a^{2}}{2}

{d o e s n}^{'} t t o u c h t h e t h r e e g i v e n c i r c l e s!

Commented by mr W last updated on 18/Jan/20

Commented by john santu last updated on 19/Jan/20

s o r r y m i s t e r, t h a t i n s t e a d i c i r c l e

t h r o u g h t h e c e n t e r o f t h r e e c i r c l e s

g i v e n .

Answered by mr W last updated on 18/Jan/20

x^{2} + y^{2} = {(2 a)}^{2} \Rightarrow c i r c l e (0, 0), r a d i u s 2 a

{(x + a)}^{2} + y^{2} = a^{2} \Rightarrow c i r c l e (- a, 0), r a d i u s a

{(x - a)}^{2} + y^{2} = a^{2} \Rightarrow c i r c l e (a, 0), r a d i u s a

f o u r t h c i r c l e (0, c), r a d i u s r

c + r = 2 a

c^{2} + a^{2} = {(a + r)}^{2}

4 a^{2} - 4 a r + r^{2} + a^{2} = a^{2} + r^{2} + 2 a r

4 a^{2} = 6 a r

\Rightarrow r = \frac{2 a}{3}

\Rightarrow c = 2 a - \frac{2 a}{3} = \frac{4 a}{3}

e q n . o f f o u r t h c i r c l e (t w o p o s s i b i l t i e s) :

x^{2} + {(y \pm \frac{4 a}{3})}^{2} = {(\frac{2 a}{3})}^{2}

Commented by mr W last updated on 18/Jan/20

Commented by peter frank last updated on 18/Jan/20

w h e r e c + r = 2 a ? c o m e f r o m s i r

Commented by mr W last updated on 18/Jan/20

Commented by mr W last updated on 18/Jan/20

O C + C B = O B

\Rightarrow c + r = 2 a

{O C}^{2} + {O A}^{2} = {A C}^{2}

\Rightarrow c^{2} + a^{2} = {(a + r)}^{2}

Commented by john santu last updated on 19/Jan/20

g o o d s i r