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10-1.Circle and System of Circles
medium
The line $(x - a)\cos \alpha + (y - b)$ $\sin \alpha = r$ will be a tangent to the circle ${(x - a)^2} + {(y - b)^2} = {r^2}$
A
If $\alpha = {30^o}$
B
If $\alpha = {60^o}$
C
For all values of $\alpha $
D
None of these
Solution
(c) According to the condition of tangency
$r = \frac{{a\cos \alpha + b\sin \alpha – (a\cos \alpha + b\sin \alpha ) – r}}{{\sqrt {{{\cos }^2}\alpha + {{\sin }^2}\alpha } }}$
$ \Rightarrow r = | – r|\; \Rightarrow r = r$.
Therefore, it is a tangent to the circle for all values of $\alpha$.
Standard 11
Mathematics
