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

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 $$\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =m+n$$Maths

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