The centre of the circle touching the circles | vedclass

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Question

(A) Let $C_1$ be $x^2+y^2-4x-6y-12=0$. Its centre $O_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let $C_2$ be $x^2+y^2+6x

Vedclass · Accepted Answer

$\left(\frac{1}{3}, -1\right)$

Vedclass · Answer

(A) Let $C_1$ be $x^2+y^2-4x-6y-12=0$. Its centre $O_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let $C_2$ be $x^2+y^2+6x+18y+26=0$. Its centre $O_2 = (-3, -9)$ and radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26} = 8$. The distance between centres $O_1O_2 = \sqrt{(2-(-3))^2 + (3-(-9))^2} = \sqrt{5^2+12^2} = 13$. Since $r_1+r_2 = 5+8 = 13 = O_1O_2$,the circles touch externally. The point of contact $P$ divides $O_1O_2$ in the ratio $r_1:r_2 = 5:8$ internally. $P = \left(\frac{5(-3)+8(2)}{5+8}, \frac{5(-9)+8(3)}{5+8}\right) = \left(\frac{-15+16}{13}, \frac{-45+24}{13}\right) = \left(\frac{1}{13}, -\frac{21}{13}\right)$. The required circle passes through $P\left(\frac{1}{13}, -\frac{21}{13}\right)$ and $A(1, -1)$. The centre $(h, k)$ lies on the line joining $O_1$ and $O_2$,which is $y-3 = \frac{-9-3}{-3-2}(x-2)$ $\Rightarrow y-3 = \frac{12}{5}(x-2)$ $\Rightarrow 12x-5y-9=0$. Testing the options,for option $A$: $12(1/3) - 5(-1) - 9 = 4+5-9 = 0$. Thus,the centre is $\left(\frac{1}{3}, -1\right)$.

The centre of the circle touching the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ at their point of contact and passing through the point $(1, -1)$ is

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