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Question Number 47179 by peter frank last updated on 05/Nov/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18
	$eqn circle (x−α)^2 +(y−β)^2 =r^2 since it touch y axis so radius r=α (x−α)^2 +(y−β)^2 =α^2 (0−α)^2 +(3−β)^2 =α^2 β=3 (x−α)^2 +(y−3)^2 =α^2 x^2 +y^2 −8x+4y−5=0 (x^2 −2.x.4+16−16)+(y^2 +4y+4−4)−5=0 (x−4)^2 +(y+2)^2 =25 distance between centre d=(√((α−4)^2 +(3+2)^2 )) cosθ=((r_1 ^2 +r_2 ^2 −d^2 )/(2r_1 r_2 )) here θ=(π/2) 25+α^2 −{(α−4)^2 +25}=0 α^2 −α^2 +8α−16=0 α=2 so eqn is (x−2)^2 +(y−3)^2 =2^2$
	$e q n c i r c l e {(x - α)}^{2} + {(y - β)}^{2} = r^{2}$ $s i n c e i t t o u c h y a x i s s o$ $r a d i u s r = α$ ${(x - α)}^{2} + {(y - β)}^{2} = α^{2}$ ${(0 - α)}^{2} + {(3 - β)}^{2} = α^{2} β = 3$ ${(x - α)}^{2} + {(y - 3)}^{2} = α^{2}$ $x^{2} + y^{2} - 8 x + 4 y - 5 = 0$ $(x^{2} - 2 . x . 4 + 16 - 16) + (y^{2} + 4 y + 4 - 4) - 5 = 0$ ${(x - 4)}^{2} + {(y + 2)}^{2} = 25$ $d i s t a n c e b e t w e e n c e n t r e$ $d = \sqrt{{(α - 4)}^{2} + {(3 + 2)}^{2}}$ $c o s θ = \frac{r_{1}^{2} + r_{2}^{2} - d^{2}}{2 r_{1} r_{2}} h e r e θ = \frac{π}{2}$ $25 + α^{2} - {{(α - 4)}^{2} + 25} = 0$ $α^{2} - α^{2} + 8 α - 16 = 0$ $α = 2$ $s o e q n i s$ ${(x - 2)}^{2} + {(y - 3)}^{2} = 2^{2}$
	Commented by peter frank last updated on 06/Nov/18

	$pls sir help Qn 47185$
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