Straight Line and Pair of Straight Lines 1 Question 49
49. No tangent can be drawn from the point $(5 / 2,1)$ to the circumcircle of the triangle with vertices $(1, \sqrt{3})$, $(1,-\sqrt{3})$ and $(3, \sqrt{3})$.
(1985, 1M)
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Solution:
- Since, $(1, \sqrt{3}),(1,-\sqrt{3})$ and $(3, \sqrt{3})$ form a right angled triangle at $(1, \sqrt{3})$
$\therefore$ Equation of circumcircle taking $(3, \sqrt{3})$ and $(1,-\sqrt{3})$ as end points of diameter.
$$ \begin{aligned} & \therefore & (x-3)(x-1)+(y-\sqrt{3})(y+\sqrt{3}) & =0 \\ & \Rightarrow & x^{2}-4 x+3+y^{2}-3 & =0 \\ & \Rightarrow & x^{2}+y^{2}-4 x & =0 \end{aligned} $$
At point
$$ \frac{5}{2}, 1, S _1=\frac{25}{4}+1-10<0 $$
$\therefore$ Point $(5 / 2,1)$ lies inside the circle.
Hence, no tangent can be drawn.
Hence, given statement is true.
