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  • The equation of a tangent to the circle passing through <img alt="(-2,11)"

The equation of a tangent to the circle \dot{x}^2+y^2=25 passing through (-2,11) is

Option: 1

4 x+3 y=25


Option: 2

3 x+4 y=38


Option: 3

24 x-7 y+125=0


Option: 4

7 x+24 y=230


Answers (1)

best_answer

The equation of tangent in terms of slope of x^2+y^2=25 is 

y=m x \pm 5 \sqrt{\left(1+m^2\right)}

Given (i), pass through (-2,11), then 

11=-2 m \pm 5 \sqrt{\left(1+m^2\right)}

Squaring both sides, then we get

\begin{aligned} & 21 m^2-44 m-96=0 \\ \\\Rightarrow \quad & (7 m-24)(3 m+4)=0 \end{aligned}

\therefore m=-4 / 3,24 / 7

There from Eq. (i) we get required tangents are 

24 x-7 y \pm 125=0 \text { and } 4 x+3 y= \pm 25

Hence tangents are 

24 x-7 y+125=0 \text { and } 4 x+3 y=25

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Gunjita

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