Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 145575 by mnjuly1970 last updated on 06/Jul/21

	Answered by ajfour last updated on 06/Jul/21

	Commented by ajfour last updated on 06/Jul/21

	$L e t t h e i n t e r s e c t i o n b e t h e$ $o r i g i n .$ $e q . o f c i r c l e$ ${(x - 2)}^{2} + y^{2} = 36$ $y = \pm \sqrt{3} x$ $\Rightarrow 4 x^{2} - 4 x - 32 = 0$ $x^{2} - x - 8 = 0$ $x_{A}, x_{B} = \frac{1 \pm \sqrt{33}}{2}$ ${A B}^{2} = {(x_{A} - x_{B})}^{2} + {(y_{A} - y_{B})}^{2}$ $= {(x_{A} - x_{B})}^{2} + 3 {(x_{A} + x_{B})}^{2}$ $= 33 + 3 = 36$ $A B = 6 (= r a d i u s)$
	Commented by ajfour last updated on 06/Jul/21
	please understand the mistake in labeling the equation of the red and blue lines.
	Commented by mnjuly1970 last updated on 06/Jul/21

	$g r a t e f u l m r a j f o r . . .$
	Answered by mr W last updated on 06/Jul/21

	Commented by mr W last updated on 06/Jul/21

	${(\frac{\sqrt{3} d}{2})}^{2} = (a + \frac{d}{2}) (b - \frac{d}{2})$ $2 d^{2} - (b - a) d - 2 a b = 0$ $d = \frac{(b - a) + \sqrt{{(b - a)}^{2} + 16 a b}}{4}$ $s i m i l a r l y$ $c = \frac{- (b - a) + \sqrt{{(b - a)}^{2} + 16 a b}}{4}$ $x^{2} = c^{2} + d^{2} - 2 c d \cos 60 °$ $= {(c + d)}^{2} - 3 c d$ $= \frac{{(b - a)}^{2} + 16 a b}{4} - 3 a b$ $= {(\frac{a + b}{2})}^{2}$ $\Rightarrow x = \frac{a + b}{2} = r a d i u s = \frac{4 + 8}{2} = 6$
	Commented by mnjuly1970 last updated on 07/Jul/21

	$e x c e l l e n t,^{″} m r W^{″} a s a l w a y s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum