If the circles given by S equiv x^2+y^2-14x+6 | vedclass

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(B) Given circles are $S_1 \equiv x^2+y^2-14x+6y+33=0$ and $S_2 \equiv x^2+y^2-a^2=0$. For two circles to have $4$ common tangents,they must be separa

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$2$

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(B) Given circles are $S_1 \equiv x^2+y^2-14x+6y+33=0$ and $S_2 \equiv x^2+y^2-a^2=0$. For two circles to have $4$ common tangents,they must be separated,which implies the distance between their centers $d > r_1 + r_2$. Center of $S_1$ is $C_1 = (7, -3)$ and radius $r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49 + 9 - 33} = \sqrt{25} = 5$. Center of $S_2$ is $C_2 = (0, 0)$ and radius $r_2 = a$. The distance between centers $d = \sqrt{(7-0)^2 + (-3-0)^2} = \sqrt{49 + 9} = \sqrt{58}$. Condition: $r_1 + r_2 Since $\sqrt{58} \approx 7.61$,we have $5 + a Since $a \in N$ (natural numbers),the possible values for $a$ are $1$ and $2$. Thus,there are $2$ possible circles.

If the circles given by $S \equiv x^2+y^2-14x+6y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0$ $(a \in N)$ have $4$ common tangents,then the possible number of circles $S^{\prime}=0$ is

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