The equations of the tangents to the circle 5 | vedclass

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(C) The given circle is $5x^2 + 5y^2 = 1$,which can be written as $x^2 + y^2 = \frac{1}{5}$.Comparing this with $x^2 + y^2 = r^2$,we get $r^2 = \frac{

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$3x + 4y = \pm \sqrt{5}$

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(C) The given circle is $5x^2 + 5y^2 = 1$,which can be written as $x^2 + y^2 = \frac{1}{5}$.Comparing this with $x^2 + y^2 = r^2$,we get $r^2 = \frac{1}{5}$,so $r = \frac{1}{\sqrt{5}}$.The line parallel to $3x + 4y = 1$ is of the form $3x + 4y + k = 0$.The distance from the center $(0, 0)$ to the tangent line is equal to the radius $r$.Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|k|}{\sqrt{3^2 + 4^2}} = \frac{1}{\sqrt{5}}$.$\frac{|k|}{5} = \frac{1}{\sqrt{5}} \Rightarrow |k| = \frac{5}{\sqrt{5}} = \sqrt{5}$.Thus,$k = \pm \sqrt{5}$.The equations of the tangents are $3x + 4y = \pm \sqrt{5}$.

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

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