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Q.
The equations of the tangents to circle $ 5x^{2}+5y^{2}=1, $ parallel to line $ 3x+4y=1 $ are
J & K CETJ & K CET 2005
Solution:
Given equation of circle is $5 x^{2}+5 y^{2}=1$
or $x^{2}+y^{2}=\frac{1}{5}$
centre of the circle is $(0,0)$.
Equation of tangent which are parallel to
$3 x+4 y-1=0$ is $3 x+4 y+\lambda=0$..(i)
We know that perpendicular distance from centre
$(0,0)$ to $3 x+4 y+\lambda=0$
should be equal to radius.
$\therefore \frac{3 \times 0+4 \times 0+\lambda}{\sqrt{(3)^{2}+(4)^{2}}}=\pm \frac{1}{\sqrt{5}}$
$\Rightarrow \lambda=\pm \frac{5}{\sqrt{5}}=\pm \sqrt{5}$
On putting the value of $\lambda$ in Eq. (i),
we get $3 x+4 y \pm \sqrt{5}=0$
or $3 x+4 y=\pm \sqrt{5}$