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10-1.Circle and System of Circles
hard
If $OA$ and $OB$ be the tangents to the circle ${x^2} + {y^2} - 6x - 8y + 21 = 0$ drawn from the origin $O$, then $AB =$
A
$11$
B
$\frac{4}{5}\sqrt {21} $
C
$\sqrt {\frac{{17}}{3}} $
D
None of these
Solution
(b) Here the equation of $AB$ (chord of contact) is
$0 + 0 – 3(x + 0) – 4(y + 0) + 21 = 0$
$ \Rightarrow 3x + 4y – 21 = 0$….$(i)$
$CM$ = perpendicular distance from $(3, 4)$ to line $(i)$ is
$\frac{{3 \times 3 + 4 \times 4 – 21}}{{\sqrt {9 + 16} }} = \frac{4}{5}$
$AM = \sqrt {A{C^2} – C{M^2}} = \sqrt {4 – \frac{{16}}{{25}}} = \frac{2}{5}\sqrt {21} $
$\therefore \;\;AB = 2AM $
$= \frac{4}{5}\sqrt {21} $ .
Standard 11
Mathematics
