Consider the equation of circles

S_1 : x^2 | vedclass

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Question

$(1)$ ${S_1} \equiv {x^2} + {y^2} + 24x - 10y + a = 0$

for real circle, $g^{2}+f^{2}-c \geq 0$

$144+25-a \geq 0$

$a \leq 169$

Also $a \geq

Vedclass · Accepted Answer

If $a = 0$, then number off common tangents to the circles $S_1 = 0$. and $S_2 = 0$ are $3$

Vedclass · Answer

$(1)$ ${S_1} \equiv {x^2} + {y^2} + 24x - 10y + a = 0$ for real circle, $g^{2}+f^{2}-c \geq 0$ $144+25-a \geq 0$ $a \leq 169$ Also $a \geq 0$ $ herefore \quad$ Total non-negative integral values of $a=170$ $(2)$ for no point in common $c_{1} c_{2}>r_{1}+r_{2}$ ${\&\,\, c_{1} c_{2}<\left|r_{1}-r_{2} ight|} $ ${c_{1} c_{2}=13} $ ${13>\sqrt{169-a}+6} $ ${\Rightarrow \quad 169-a<49} $ $a > 120$ and $a \le 169$ So in this condition we have $49$ integral values of a But from $c_{1} c_{2}<\left|r_{1}-r_{2} ight|,$ we will get additional values of a. So $B$ is false $(3)$ for orthogonal cut ${2.12 .0+2(-5) .0=-36+a} $ ${\Rightarrow \quad a=36}$ $(4)$ If $a=0, c_{1} c_{2}=13 $ and $ r_{1}+r_{2}=19$ ${c_1}{c_2} < {r_1} + {r_2}$ No. of common tangent $=2$

Consider the equation of circles

$S_1 : x^2 + y^2 + 24x - 10y + a = 0$

$S_2 : x^2 + y^2 = 36$ which of the following is not correct

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