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Question

A line meets the coordinate axes in $$A$$ and $$B$$. A circle is circumscribed about the $$\triangle AOB$$. If $$m,n$$ are the distance of the tangent to the circle at the origin from the points $$A$$ and $$B$$, respectively the diameter of the circle is


A
m(m+n)
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B
m+n
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C
n(m+n)
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D
None of these
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Solution

The correct option is A $$m+n$$
Let $$A\equiv \left( a,0 \right) $$ and $$B\equiv \left( 0,b \right) $$
Since $$\angle AOB = {90}^{0}$$, $$\therefore AB$$ is the diameter
$$\therefore$$ Center of the circle is $$\displaystyle$$ and radius $$\displaystyle =\frac { 1 }{ 2 } \sqrt { { a }^{ 2 }+{ b }^{ 2 } } $$.
$$\therefore$$ Equation of the circle is 
$$\displaystyle { \left( x-\frac { a }{ 2 }  \right)  }^{ 2 }+{ \left( y-\frac { b }{ 2 }  \right)  }^{ 2 }=\frac { 1 }{ 4 } \left( { a }^{ 2 }+{ b }^{ 2 } \right) \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-ax-by=0$$
Equation of tangent to the circle at $$O(0,0)$$ is $$ax+by=0$$   ...(1)
$$\therefore m=$$ length of $$\bot$$ from $$A(a,0)$$ on (1) $$\displaystyle =\frac { { a }^{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  } $$
and $$n=$$ length $$\bot$$ from $$(0,b)$$ on (1) $$\displaystyle =\frac { { b }^{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  } $$
$$\therefore$$ Diameter $$\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =m+n$$

389325_257316_ans_2549948c6d8d402dba0d146592263bed.png

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