Find-the-equation-to-two-circles-which-touch-the-x-axis-at-the-origin-and-also-touch-the-line-12x-5y-60-

Question Number 54270 by peter frank last updated on 01/Feb/19

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

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

x - a x i s a t t h e o r i g i n

a n d a l s o t o u c h t h e l i n e

12 x + 5 y = 60

Answered by ajfour last updated on 01/Feb/19

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

c = k \pm k \sqrt{1 + m^{2}} \Rightarrow 12 = k \pm k \sqrt{1 + {(\frac{12}{5})}^{2}}

\Rightarrow k = \frac{12}{1 \pm \frac{13}{5}} = \frac{10}{3}, - \frac{15}{2}

h e n c e e q . o f c i r c l e a b o v e x - a x i s i s

3 x^{2} + 3 y^{2} - 20 y = 0

w h i l e t h a t b e l o w x - a x i s i s

x^{2} + y^{2} + 15 y = 0 .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

x^{2} + {(y - a)}^{2} = a^{2} c e n t r e (0, a)

∣ \frac{12 \times 0 + 5 a - 60}{\sqrt{12^{2} + 5^{2}}} ∣= a

5 a - 60 = 13 a

a = \frac{- 60}{8} = \frac{- 15}{2}

x^{2} + {(y + 7.5)}^{2} = {(7.5)}^{2}

e d i t e d f o r s e c o n d e q u a t i i n c e n t r e (0, - a)

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

∣ \frac{12 \times 0 + 5 (- a) - 60}{\sqrt{12^{2} + 5^{2}}} ∣= a

- 5 a - 60 = 13 a

18 a = - 60 a = \frac{- 10}{3}

x^{2} + {(y - \frac{10}{3})}^{2} = \frac{100}{9}

x^{2} + y^{2} - \frac{20 y}{3} = 0

3 x^{2} + 3 y^{2} - 20 y = 0