The equation of the circle whose radius is 3 | vedclass

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(B) The equation of the given circle is $x^2+y^2-4x-6y-12=0$. Its center is $C_1(2,3)$ and radius is $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$.

Vedclass · Accepted Answer

$5x^2+5y^2-8x-14y-32=0$

Vedclass · Answer

(B) The equation of the given circle is $x^2+y^2-4x-6y-12=0$. Its center is $C_1(2,3)$ and radius is $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$.Let the required circle have center $C_2(h, k)$ and radius $r_2 = 3$. It touches the given circle internally at $A(-1,-1)$.The point $A$ divides the line segment $C_1C_2$ externally in the ratio $r_1:r_2 = 5:3$.Using the section formula for external division:$(-1, -1) = \left( \frac{5h - 3(2)}{5-3}, \frac{5k - 3(3)}{5-3} \right)$$(-1, -1) = \left( \frac{5h-6}{2}, \frac{5k-9}{2} \right)$Equating coordinates:$5h-6 = -2$ $\Rightarrow 5h = 4$ $\Rightarrow h = \frac{4}{5}$$5k-9 = -2$ $\Rightarrow 5k = 7$ $\Rightarrow k = \frac{7}{5}$The equation of the required circle is $(x-h)^2 + (y-k)^2 = r_2^2$:$(x-\frac{4}{5})^2 + (y-\frac{7}{5})^2 = 3^2$$x^2 - \frac{8x}{5} + \frac{16}{25} + y^2 - \frac{14y}{5} + \frac{49}{25} = 9$$x^2 + y^2 - \frac{8x}{5} - \frac{14y}{5} + \frac{65}{25} = 9$$x^2 + y^2 - \frac{8x}{5} - \frac{14y}{5} + \frac{13}{5} = 9$Multiplying by $5$:$5x^2 + 5y^2 - 8x - 14y + 13 = 45$$5x^2 + 5y^2 - 8x - 14y - 32 = 0$Thus,option $B$ is correct.

The equation of the circle whose radius is $3$ and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is

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