The equations of the tangent to the circle $5 x^2+5 y^2=1$ parallel to the line $3 x+4 y=1$ are

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Aicharya · Accepted Answer

Given, equation of circle is $5 x^2+5 y^2=1$

$\begin{aligned} & \Rightarrow \quad x^2+y^2=(1 / \sqrt{5})^2 \\ & \therefore \text { Centre }=(0,0) \text { and radius }=1 / \sqrt{5}\end{aligned}$ Given, line is 3x + 4y = 1 $ \Rightarrow \text { Slope }(m)=\frac{-3}{4} $ Since, we know that tangents to the circle $x^2+y^2=a^2$ in the slope form is given by $ y=m x \pm a \sqrt{1+m^2} ...(i) $ Since, required line is parallel to given line, hence it will have same slope and radius $=\frac{1}{\sqrt{5}}$ From Eq. (i), we get $\begin{aligned} & y=\frac{-3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}=\frac{-3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}} \\ & y=\frac{-3 x}{4} \pm \frac{1}{\sqrt{5}} \times \frac{5}{4} \Rightarrow 4 y=-3 x \pm \sqrt{5} \\ & \Rightarrow \quad 3 x+4 y= \pm \sqrt{5}\end{aligned}$

The equations of the tangent to the circle $5 x^2+5 y^2=1$ parallel to the line $3 x+4 y=1$ are

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