The equations of three circles are given
$x^{2} + y^{2} = 1, x^{2} + y^{2} + 8 x + 15 = 0, x^{2} + y^{2} + 10 y + 24 = 0$ .
Determine the co-ordinates of the point such that the tangents drawn from it to three circles are equal in length.

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Solution

Let the point be $(h, k)$ then by the given condition $t^{2} = h^{2} + k^{2} - 1 = h^{2} + k^{2} + 8 h + 15$
$= h^{2} + k^{2} + 10 k + 24$ .
From $1 s t$ and $2 n d$ , we get $h = - 2$
From $1 s t$ and $3 r d$ , we get $k = - 5 / 2$ .
Hence the required point is
$(- 2, - 5 / 2)$ and $t^{2} = 37 / 4$ .

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The equations of three circles are givenx2+y2=1,x2+y2+8x+15=0,x2+y2+10y+24=0.Determine the co-ordinates of the point such that the tangents drawn from it to three circles are equal in length.

Let the point be (h,k) then by the given condition t2=h2+k2−1=h2+k2+8h+15=h2+k2+10k+24.From 1st and 2nd, we get h=−2From 1st and 3rd, we get k=−5/2.Hence the required point is(−2,−5/2) and t2=37/4.

The equations of three circles are given
$x^{2} + y^{2} = 1, x^{2} + y^{2} + 8 x + 15 = 0, x^{2} + y^{2} + 10 y + 24 = 0$ .
Determine the co-ordinates of the point such that the tangents drawn from it to three circles are equal in length.