Given the three circles $x^{2} + y^{2} - 16 x + 60 = 0, 3 x^{2} + 3 y^{2} - 36 x + 81 = 0$ , and $x^{2} + y^{2} - 16 x - 12 y + 84 = 0$ , find (1) the point from which the tangents to them are equal in length, and (2) this length.

Open in App

Solution

Equations of given circles are

$x^{2} + y^{2} - 16 x + 60 = 0............ (i) 3 x^{2} + 3 y^{2} - 36 x + 81 = 0 \Rightarrow x^{2} + y^{2} - 12 x + 27 = 0.......... (i i) x^{2} + y^{2} - 16 x + 12 y + 84 = 0....... (i i i)$

Let tangents are drawn from $P (h, k)$ and it touches the circles $(i), (i i)$ and $(i i i)$ at $T_{1}, T_{2}$ and $T_{3}$ repectively

Given $P T_{1} = P T_{2} = P T_{3}$

$\Rightarrow \sqrt{h^{2} + k^{2} - 16 h + 60} = \sqrt{h^{2} + k^{2} - 12 h + 27} \Rightarrow h^{2} + k^{2} - 16 h + 60 = h^{2} + k^{2} - 12 h + 27 \Rightarrow h = \frac{33}{4}$

Also $\sqrt{h^{2} + k^{2} - 16 h + 60} = \sqrt{h^{2} + k^{2} - 16 h + 12 k + 84}$

$\Rightarrow h^{2} + k^{2} - 16 h + 60 = h^{2} + k^{2} - 16 h + 12 k + 84 \Rightarrow k = - 2$

So the points from which equal tangents can be drawn is $(\frac{33}{4}, - 2)$

Let length of tangent be $l$

Now $l = {P T}_{1}$

$\Rightarrow l = \sqrt{h^{2} + k^{2} - 16 h + 60} \Rightarrow l = \sqrt{{(\frac{33}{4})}^{2} + {(- 2)}^{2} - 16 (\frac{33}{4}) + 60} \Rightarrow l = \sqrt{\frac{1}{16}} = \frac{1}{4}$

Suggest Corrections

Similar questions

S_{1} : x^{2} + y^{2} - 16 x + 60 = 0,

S_{2} : x^{2} + y^{2} - 12 x + 20 = 0,

S_{3} : x^{2} + y^{2} - 16 x - 12 y = 0

L

L

x^{2} + y^{2} - 4 = 0, x^{2} + y^{2}

+ 3 y = 0

x^{2} + y^{2} + 7 y - 18 = 0

x^{2} + y^{2} + 4 x + 6 y - 19 = 0

x^{2} + y^{2} - 2 x - 2 y - 5 = 0

x^{2} + y^{2} - 9 = 0

x^{2} + y^{2} - 2 x + 3 y - 7 = 0

x^{2} + y^{2} + 5 x - 5 y + 9 = 0

x^{2} + y^{2} + 7 x - 9 y + 29 = 0

x^{2} + y^{2} = 1, x^{2} + y^{2} + 8 x + 15 = 0, x^{2} + y^{2} + 10 y + 24 = 0

Join BYJU'S Learning Program