-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="AP EAPCET 2021 - 20th August Morning Shift Mathematics - Circle Question 143 English Explanation">

Let $$A = (3\cos \theta ,3\sin \theta )$$

Then, $$B = \left( {3\cos \left( {\theta + {{2\pi } \over 3}} \right),3\sin \left( {\theta + {{2\pi } \over 3}} \right)} \right)$$

Let (h, k) be the required point

$$2h = 3\cos \theta + 3\cos \left( {\theta + {{2\pi } \over 3}} \right)$$

$$ = 3\left[ {\cos \theta + \cos \theta \left( { - {1 \over 2}} \right) - \sin \theta \left( {{{\sqrt 3 } \over 2}} \right)} \right]$$

$$ = {3 \over 2}(\cos \theta - \sqrt 3 \sin \theta )$$

$$2k = 3\sin \theta + 3\sin \left( {\theta + {{2\pi } \over 3}} \right)$$

$$ = 3\left[ {\sin \theta + \sin \theta \left( {{{ - 1} \over 3}} \right) + \cos \theta \left( {{{\sqrt 3 } \over 2}} \right)} \right]$$

$$ = {3 \over 2}(\sin \theta + \sqrt 3 \cos \theta )$$

$$\cos \theta - \sqrt 3 \sin \theta = {{4h} \over 3} \Rightarrow \sqrt 3 \cos \theta + \sin \theta = {{4k} \over 3}$$

$$\sqrt 3 \cos \theta - 3\sin \theta = {{4h} \over {\sqrt 3 }}$$

$$4\sin \theta = {4 \over 3}(k - \sqrt 3 h)$$

$$\sin \theta = \left( {{{k - \sqrt 3 h} \over 3}} \right)$$

$$\cos \theta = \left( {{{3k + \sqrt 3 h} \over {3\sqrt 3 }}} \right) = \left( {{{\sqrt 3 k + h} \over 3}} \right)$$

$$ \Rightarrow {\left( {{{k - \sqrt 3 h} \over 3}} \right)^2} + {\left( {{{\sqrt 3 k + h} \over 3}} \right)^2} = 1$$

$$ \Rightarrow {k^2} + 3{h^2} - 2\sqrt 3 hk + 3{k^2} + {h^2} + 2\sqrt 3 kh = 9$$

$$ \Rightarrow 4({k^2} + {h^2}) = 9 \Rightarrow {x^2} + {y^2} = {9 \over 4}$$

Circle

Continue this paper

Similar Circle PYQs

More practice from this question