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(D) The circle $S$ lies in the first quadrant and touches both coordinate axes with radius $r_1 = 2$. Thus,its centre is $C_1 = (2, 2)$.Let the requir

Vedclass · Accepted Answer

$x^2+y^2-12x-10y+52=0$

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(D) The circle $S$ lies in the first quadrant and touches both coordinate axes with radius $r_1 = 2$. Thus,its centre is $C_1 = (2, 2)$.Let the required circle have centre $C_2 = (6, 5)$ and radius $r$.Since the two circles touch each other externally,the distance between their centres is equal to the sum of their radii:$C_1C_2 = r_1 + r$Using the distance formula,$C_1C_2 = \sqrt{(6-2)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$.So,$5 = 2 + r$,which gives $r = 3$.The equation of the circle with centre $(6, 5)$ and radius $3$ is:$(x-6)^2 + (y-5)^2 = 3^2$$x^2 - 12x + 36 + y^2 - 10y + 25 = 9$$x^2 + y^2 - 12x - 10y + 61 - 9 = 0$$x^2 + y^2 - 12x - 10y + 52 = 0$Thus,the correct option is $D$.

$A$ circle $S$ of radius $2$ units lies in the first quadrant and touches both the coordinate axes. The equation of the circle with centre at $(6,5)$ and touching the circle $S$ externally is

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