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10-1.Circle and System of Circles
easy
રેખા $4x + 3y + 5 = 0$ ને સમાંતર, વર્તૂળ $x^2 + y^2 - 6x + 4y = 12$ ની સ્પર્શક રેખાઓ :
A
$4 x+3 y-31=0,4 x+3 y+19=0$
B
$4 x+3 y+5=0,4 x+3 y-25=0$
C
$4x + 3y - 17 =0, 4x + 3y + 13 = 0$
D
એકપણ નહિ
Solution
$x^2+y^2-6 x+4 y=12$
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 "
$y=\frac{-4 x}{3} \pm 4 \sqrt{1+\frac{16}{9}}$
$y=\frac{-4 x}{3} \pm 5 \times \frac{5}{3}$
$3 y =-4 x \pm 25 \Rightarrow$ so here $3 y +4 x +25=0$
or $3 y+4 x-25=0$
Standard 11
Mathematics
