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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
Number of non-negative integral values of $'a'$ such that $S_1 = 0$ represents a real circle $170$
If $S_1 = 0$ and $S_2 = 0$ has no point in common, then number of integral values of $a$ is more than $49$
If $S_1 = 0$ and $S_2 = 0$ intersect orthogonally then $a = 36$
If $a = 0$, then number off common tangents to the circles $S_1 = 0$. and $S_2 = 0$ are $3$
Solution
$(1)$ ${S_1} \equiv {x^2} + {y^2} + 24x – 10y + a = 0$
for real circle, $g^{2}+f^{2}-c \geq 0$
$144+25-a \geq 0$
$a \leq 169$
Also $a \geq 0$
$\therefore \quad$ 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}\right|} $
${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 $c_{1} c_{2}<\left|r_{1}-r_{2}\right|,$
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 $=2$
