રેખા 4x + 3y + 5 = 0 ને સમાંતર, વર્તૂળ x^2 + | vedclass

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Question

$x^2+y^2-6 x+4 y=12$

Centre of circle $=(-g,-f)=(3,-2)$

and radius $a=\sqrt{g^2+f^2-c}=\sqrt{9+14+12}$ $=5$

So $eq ^{ n }$ of tangent $y = mx

Vedclass · Accepted Answer

$4 x+3 y+5=0,4 x+3 y-25=0$

Vedclass · Answer

$x^2+y^2-6 x+4 y=12$

Centre of circle $=(-g,-f)=(3,-2)$

and radius $a=\sqrt{g^2+f^2-c}=\sqrt{9+14+12}$ $=5$

So $eq ^{ n }$ of tangent $y = mx \pm \sqrt{1+ m ^2}$

So $y = mx \pm 5 \sqrt{1+ m ^2} \ldots$. (1)

but here tangent is parallel to the line $4 x+3 y+5=0$ $3 y =-5-4 x$

$y=\frac{-5}{3}-\frac{4}{3} x$

So here slope $=\frac{-4}{3}$

So from (i) tangent eq "

$y=\frac{-4 x}{3} \pm 4 \sqrt{1+\frac{16}{9}}$

$y=\frac{-4 x}{3} \pm 5 	imes \frac{5}{3}$

$3 y =-4 x \pm 25 \Rightarrow$ so here $3 y +4 x +25=0$

or $3 y+4 x-25=0$

રેખા $4x + 3y + 5 = 0$ ને સમાંતર, વર્તૂળ $x^2 + y^2 - 6x + 4y = 12$ ની સ્પર્શક રેખાઓ :

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