Lines 5x+12y−10=0 and 5x−12y−40=0 touch circle C_1 of diameter 6. If the center of C_1 lies in the Ist quadrant, find the equation of circle C_2 which is concentric with C_(1 ) and cuts intercept of length 8 on these lines.


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 62698 by Ankit0512 last updated on 24/Jun/19

$L i n e s 5 x + 12 y - 10 = 0 a n d 5 x - 12 y - 40 = 0$ $t o u c h c i r c l e C_{1} o f d i a m e t e r 6 . I f t h e$ $c e n t e r o f C_{1} l i e s i n t h e I s t q u a d r a n t,$ $f i n d t h e e q u a t i o n o f c i r c l e C_{2} w h i c h i s$ $c o n c e n t r i c w i t h C_{1} a n d c u t s i n t e r c e p t$ $o f l e n g t h 8 o n t h e s e l i n e s .$
	Answered by tanmay last updated on 24/Jun/19

	$C_{1} {(x - a)}^{2} + {(y - b)}^{2} = 3^{2} [r = \frac{d}{2} = \frac{6}{2} = 3]$ $∣ \frac{5 a + 12 b - 10}{\sqrt{5^{2} + 12^{2}}} ∣=∣ \frac{5 a - 12 b - 40}{\sqrt{5^{2} + {(- 12)}^{2}}} ∣= 3$ $5 a + 12 b - 10 = 39$ $5 a - 12 b - 40 = 39$ $10 a = 78 + 50 \to a = 12.8$ $24 b + 30 = 0 \to b = - \frac{5}{4}$ $C_{1} {(x - 12.8)}^{2} + {(y + \frac{5}{4})}^{2} = 3^{2}$ $C_{2} {(x - 12.8)}^{2} + {(y + \frac{5}{4})}^{2} = R^{2}$ $d i s t a n c e f r o m (12.8, \frac{5}{4}) t o l i n e 5 x + 12 y - 10 = 0$ $∣ \frac{5 \times 12.8 - 12 \times \frac{5}{4} - 10}{\sqrt{5^{2} + 12^{2}}} ∣= h (s a y)$ $∣ \frac{39}{13} ∣= h$ $R^{2} = h^{2} + {(\frac{8}{2})}^{2}$ $R^{2} = 9 + 16 = 25$ $C_{2} {(x - 12.8)}^{2} + {(y + \frac{5}{4})}^{2} = 25$
	Commented by ajfour last updated on 24/Jun/19

	$p l e a s e c h e c k, S i r . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum