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(B) The equation of a circle with radius $r$ touching the positive coordinate axes is $(x-r)^2+(y-r)^2=r^2$. Given circle: $x^2+y^2-12x-10y+52=0$. Rew

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The equation of a circle with radius $r$ touching the positive coordinate axes is $(x-r)^2+(y-r)^2=r^2$. Given circle: $x^2+y^2-12x-10y+52=0$. Rewriting the given circle as $(x-6)^2+(y-5)^2 = 36+25-52 = 9$,so the centre $C_2 = (6, 5)$ and radius $r_2 = 3$. Since the circles touch externally,the distance between centres $C_1(r, r)$ and $C_2(6, 5)$ is $d = r_1+r_2 = r+3$. Thus,$\sqrt{(r-6)^2+(r-5)^2} = r+3$. Squaring both sides: $(r-6)^2+(r-5)^2 = (r+3)^2$. $r^2-12r+36+r^2-10r+25 = r^2+6r+9$. $r^2-28r+52 = 0$. $(r-2)(r-26) = 0$,so $r=2$ or $r=26$. For $r=2$,the distance between centres is $r+3 = 2+3 = 5$. For $r=26$,the distance between centres is $r+3 = 26+3 = 29$. Since $5$ is one of the options,the correct answer is $5$.

If a circle of radius $r$ touches the positive coordinate axes and also the circle $x^2+y^2-12x-10y+52=0$ externally,then the distance between the centres of the two circles is

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