- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
medium
वृत्त ${x^2} + {y^2} = {a^2}$ पर बिन्दु $(\alpha ,\beta )$ से खींची गयी स्पर्श रेखाओं के बीच कोण है
A
${\tan ^{ - 1}}\left( {\frac{a}{{\sqrt {{\alpha ^2} + {\beta ^2} - {a^2}} }}} \right)$
B
${\tan ^{ - 1}}\left( {\frac{{\sqrt {{\alpha ^2} + {\beta ^2} - {a^2}} }}{a}} \right)$
C
$2{\tan ^{ - 1}}\left( {\frac{a}{{\sqrt {{\alpha ^2} + {\beta ^2} - {a^2}} }}} \right)$
D
इनमें से कोई नहीं
Solution
(c) $\tan \frac{\theta }{2} = \frac{{C{T_1}}}{{P{T_1}}} $
$= \frac{a}{{\sqrt {{\alpha ^2} + {\beta ^2} – {a^2}} }}$
$\frac{\theta }{2} = {\tan ^{ – 1}}\frac{a}{{\sqrt {{\alpha ^2} + {\beta ^2} – {a^2}} }} $
$\Rightarrow \theta = 2{\tan ^{ – 1}}\frac{a}{{\sqrt {{\alpha ^2} + {\beta ^2} – {a^2}} }}$.
Standard 11
Mathematics
