The equation of the circle whose radius is 5 | vedclass

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(B) The given circle is ${x^2} + {y^2} - 2x - 4y - 20 = 0$. Its center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-20)} = \sqrt{25} = 5$. Let

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${x^2} + {y^2} - 18x - 16y + 120 = 0$

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(B) The given circle is ${x^2} + {y^2} - 2x - 4y - 20 = 0$. Its center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-20)} = \sqrt{25} = 5$. Let the required circle have center $C_2 = (h, k)$ and radius $r_2 = 5$. Since the circles touch externally at $(5, 5)$,the point $(5, 5)$ divides the line segment joining the centers $C_1(1, 2)$ and $C_2(h, k)$ in the ratio $r_1 : r_2 = 5 : 5 = 1 : 1$. Thus,$(5, 5) = (\frac{1+h}{2}, \frac{2+k}{2})$. Solving for $h$ and $k$: $1 + h = 10 \Rightarrow h = 9$ $2 + k = 10 \Rightarrow k = 8$. The equation of the required circle is $(x - 9)^2 + (y - 8)^2 = 5^2$. Expanding this,we get ${x^2} - 18x + 81 + {y^2} - 16y + 64 = 25$. ${x^2} + {y^2} - 18x - 16y + 120 = 0$.

The equation of the circle whose radius is $5$ and which touches the circle ${x^2} + {y^2} - 2x - 4y - 20 = 0$ externally at the point $(5, 5)$ is

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