The lengths of the tangents from A, B to a circle are $6, 7$ respectively. If $A$ and $B$ are conjugate points then $A B =$

\sqrt{\frac{85}{2}}

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2 \sqrt{85}

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\sqrt{85}

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\sqrt{\frac{85}{4}}

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Solution

The correct option is C $\sqrt{85}$
By property , $A$ and $B$ are conjugate points when
$B$ lies on polar of $A$ and also $A$ lies on polar $B$ .
So, distance between $A$ and $B$ $= \sqrt{(L_{1})^{2} + (L_{2})^{2}}$
So, $L_{1} = 6$ and $L_{2} = 7$
$∴ A B = \sqrt{36 + 49} = \sqrt{85}$

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The lengths of the tangents from A, B to a circle are 6,7 respectively. If A and B are conjugate points then AB=

The correct option is C √85By property , A and B are conjugate points whenB lies on polar of A and also A lies on polar B.So, distance between A and B =√(L1)2+(L2)2So, L1=6 and L2=7∴AB=√36+49=√85

The lengths of the tangents from A, B to a circle are $6, 7$ respectively. If $A$ and $B$ are conjugate points then $A B =$

The correct option is C $\sqrt{85}$
By property , $A$ and $B$ are conjugate points when
$B$ lies on polar of $A$ and also $A$ lies on polar $B$ .
So, distance between $A$ and $B$ $= \sqrt{(L_{1})^{2} + (L_{2})^{2}}$
So, $L_{1} = 6$ and $L_{2} = 7$
$∴ A B = \sqrt{36 + 49} = \sqrt{85}$