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From a point A (4, 3) rays are drawn to the circle \mathrm{45 x-28 y+106=0}. If the length of ray AB is the longest and C is the centre of the circle, then the area of triangle ACB is

Option: 1

144


Option: 2

72


Option: 3

36


Option: 4

34


Answers (1)

best_answer

The longest ray from A is along a tangential direction to the circle from A. Therefore, AB is

a tangent to the circle.

\mathrm{\text { Length } \begin{aligned} A B & =\sqrt{S_1} \\ & =\sqrt{16+9+64+54+1}=\sqrt{144}=12 \end{aligned}}

As the centre of the circle is (–8, –9),

its radius \mathrm{=\sqrt{A C^2-A B^2}=\sqrt{144+144-144}=\sqrt{144}=12}

Area of right-angled 

\mathrm{\begin{aligned} \triangle C B A & =\frac{1}{2} C B \cdot B A \\ & =\frac{1}{2}(12)^2=72 \end{aligned}}

 

 

Posted by

Ritika Harsh

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