Coordinates of the centre will be: \[ (2, 1) \]
Equation of circle: \[ (x - 2)^2 + (y - 1)^2 = 1 \]
\[ QC = \sqrt{(5 - 2)^2 + (5 - 1)^2} = 5 \]
Shortest distance: \[ RQ = CQ - CR = 5 - 1 = 4 \]
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB.
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: