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10-1.Circle and System of Circles
normal
Tangents are drawn from $(4, 4) $ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ to meet the circle at $A$ and $B$. The length of the chord $AB $ is
A
$2\sqrt{3}$
B
$3\sqrt{2}$
C
$2\sqrt{6}$
D
$6\sqrt{2}$
Solution
length of $C.O.C. = \frac{{2RL}}{{\sqrt {{R^2} + {L^2}} }}$
equation of circle $x^2 + y^2 – 2x – 2y – 7 = 0$
equation of $COC = 4x + 4y – (x + 4) – (y + 4) – 7 = 0 = 3x + 3y – 15 = 0 $
radius $= 3$ perpendicular from $(1, 1)$ =$\left| {\,\frac{{6 – 15}}{{\sqrt {18} }}\,} \right|$ =$\frac{9}{{\sqrt {18} }}$ =$\frac{9}{{3\sqrt 2 }}$ =$\frac{3}{{\sqrt 2 }}$
$\therefore$ length $\frac{{AB}}{2}$ = $\sqrt {{r^2} – {{\left( {\frac{3}{{\sqrt 2 }}} \right)}^2}} $ = $\frac{3}{{\sqrt 2 }}$
$\therefore$ $AB = 3\sqrt{2}$
Standard 11
Mathematics
