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  • Given a circle with equation x^2 + y^2 = 1 and a line with equation y = mx + c. Find the point of intersection of the tangents drawn to the circle from the point of intersection of the li

Given a circle with equation x^2 + y^2 = 1 and a line with equation y = mx + c. Find the point of intersection of the tangents drawn to the circle from the point of intersection of the line and the circle.

Option: 1

\mathrm{\left(x_1, x_1 \sqrt{m^2}\right) and \left(x_1,-x_1 \sqrt{m^2}\right)}
 


Option: 2

\mathrm{\left(x_1,-x_1 \sqrt{m^2}\right) \operatorname{and}\left(x_1,-x_1 \sqrt{m^2}\right)}
 


Option: 3

\mathrm{\left(x_1,-x_1\right) and \left(\sqrt{m^2},-x_1 \sqrt{m^2}\right)}
 


Option: 4

\mathrm{\left(x_1, 0\right) \operatorname{and}\left(\sqrt{m^2}, 0\right)}


Answers (1)

best_answer

Let \mathrm{(x 1, y 1)} be the point of intersection of the line and the circle.
Here, \mathrm{x_1^2+\left(m x_1+c\right)^2=1}
On expanding, \mathrm{x_1^2+m^2 x_1^2+2 m c_1+c^2=1}
On rearranging, \mathrm{\left(m^1+1\right) x_1^2+2 m c x_1+\left(c^2-1\right)=0}
This equation has two solutions, tangents to the circle from \mathrm{(x 1, y 1)} have slope \mathrm{\frac{1}{m},-\frac{1}{m}} Equations of tangents are

\mathrm{ \begin{aligned} & \Rightarrow y-y_1=-\frac{1}{m}\left(x-x_1\right) \text { and } y-y_1=\frac{1}{m}\left(x-x_1\right) \\ & \Rightarrow x=\frac{m^2 y_1+m x_1+y_1}{m^2+1} \text { and } x=\frac{m^2 y_1-m x_1+y_1}{m^2+1} \end{aligned} }

Substituting these into the equation of the circle,

\mathrm{\Rightarrow\left(\frac{m^2 y_1+m x_1+y_1}{m^2+1}\right)^2+\left(\frac{m^2 y_1-m x_1+y_1}{m^2+1}\right)^2=1 \\}

\mathrm{\Rightarrow \frac{m^4 y_1^2+2 m^3 y_1 x_1+m^2 x_1^2+m^2 y_1^2-2 m y_1 x_1+x_1^2}{m^2+1}=1 \\}

\mathrm{\Rightarrow m^4 y_1^2+m^2 x_1^2=m^2 y_1^2+x_1^2 \\}

\mathrm{m^4 y_1^2-x_1^2=0 \text {, since } m \neq 0 \text { divide both sides by } m^2 \\}

\mathrm{\Rightarrow y_1= \pm x_1 \sqrt{m^2} \\}

The points of intersection of the tangents are\mathrm{\left(x_1, x_1 \sqrt{m^2}\right) and \left(x_1,-x_1 \sqrt{m^2}\right).}

Posted by

himanshu.meshram

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