The equations of the tangent to the circle 5x | vedclass

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(B) Given,the equation of the circle is $5x^2 + 5y^2 = 1$. Dividing by $5$,we get $x^2 + y^2 = \frac{1}{5} = \left(\frac{1}{\sqrt{5}}\right)^2$. Thus,

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

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(B) Given,the equation of the circle is $5x^2 + 5y^2 = 1$. Dividing by $5$,we get $x^2 + y^2 = \frac{1}{5} = \left(\frac{1}{\sqrt{5}}\right)^2$. Thus,the center is $(0, 0)$ and the radius $r = \frac{1}{\sqrt{5}}$. The given line is $3x + 4y = 1$,which has a slope $m = -\frac{3}{4}$. The equation of a tangent to the circle $x^2 + y^2 = r^2$ with slope $m$ is given by $y = mx \pm r\sqrt{1 + m^2}$. Substituting $m = -\frac{3}{4}$ and $r = \frac{1}{\sqrt{5}}$: $y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \left(-\frac{3}{4}\right)^2}$ $y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \frac{9}{16}}$ $y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}$ $y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \cdot \frac{5}{4}$ $y = -\frac{3}{4}x \pm \frac{\sqrt{5}}{4}$ Multiplying by $4$: $4y = -3x \pm \sqrt{5}$ $3x + 4y = \pm \sqrt{5}$.

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

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