Given three collinear points A(3,1),B(7,-1),a | vedclass

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(C) For a point $A$ outside a circle,if a secant through $A$ intersects the circle at points $C$ and $B$,then the length of the tangent $AT$ from $A$

Vedclass · Accepted Answer

$\sqrt{10}$

Vedclass · Answer

(C) For a point $A$ outside a circle,if a secant through $A$ intersects the circle at points $C$ and $B$,then the length of the tangent $AT$ from $A$ to the circle is given by the Power of a Point theorem: $AT^2 = AC \cdot AB$.First,calculate the distances $AC$ and $AB$ using the distance formula:$AC = \sqrt{(5-3)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$$AB = \sqrt{(7-3)^2 + (-1-1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$Now,substitute these values into the formula:$AT^2 = AC \cdot AB = \sqrt{5} \cdot 2\sqrt{5} = 2 \cdot 5 = 10$$AT = \sqrt{10}$ units.

Given three collinear points $A(3,1)$,$B(7,-1)$,and $C(5,0)$. The length of a tangent drawn from $A$ to any circle that passes through $B$ and $C$ is ....... units.

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