A point that lies on the common tangent to th | vedclass

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Question

(D) Given circles are $S_1 \equiv x^2+y^2-2x+18y+78=0$ and $S_2 \equiv x^2+y^2+8x-6y-200=0$. Centre $C_1 = (1, -9)$ and radius $r_1 = \sqrt{1^2+(-9)^2

Vedclass · Accepted Answer

$\left(\frac{-2}{5}, \frac{-47}{4}\right)$

Vedclass · Answer

(D) Given circles are $S_1 \equiv x^2+y^2-2x+18y+78=0$ and $S_2 \equiv x^2+y^2+8x-6y-200=0$. Centre $C_1 = (1, -9)$ and radius $r_1 = \sqrt{1^2+(-9)^2-78} = \sqrt{1+81-78} = 2$. Centre $C_2 = (-4, 3)$ and radius $r_2 = \sqrt{(-4)^2+3^2-(-200)} = \sqrt{16+9+200} = 15$. Distance $C_1C_2 = \sqrt{(-4-1)^2+(3-(-9))^2} = \sqrt{(-5)^2+12^2} = \sqrt{25+144} = 13$. Since $r_2 - r_1 = 15 - 2 = 13 = C_1C_2$,the circles touch each other internally. The equation of the common tangent is $S_1 - S_2 = 0$. $(x^2+y^2-2x+18y+78) - (x^2+y^2+8x-6y-200) = 0$. $-10x + 24y + 278 = 0$,which simplifies to $-5x + 12y + 139 = 0$. Checking option $D$: $-5\left(\frac{-2}{5}\right) + 12\left(\frac{-47}{4}\right) + 139 = 2 - 141 + 139 = 0$. Thus,the point $\left(\frac{-2}{5}, \frac{-47}{4}\right)$ lies on the common tangent.

$A$ point that lies on the common tangent to the circles $x^2+y^2-2x+18y+78=0$ and $x^2+y^2+8x-6y-200=0$ among the following options is

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