MathematicsRadius of circumcircle of a triangle ABC is 510 units. If point P is equidistant from A(1,3),B(-3,5) and C(5,-1) and then find AP.

Radius of circumcircle of a triangle ABC is $5 \sqrt{10}$ units. If point P is equidistant from $A (1, 3)$ , $B (- 3, 5)$ and $C (5, - 1)$ and then find AP.

Given,
Radius of circle=

5 \sqrt{10}

Drawing the required figure,

The distance between

A (x_{1}, y_{1})

and

B (x_{2}, y_{2})

\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

.
If point P is equidistant from A, B and C, then AP is the radius of the circle.
AP

= 5 \sqrt{10}

units.
Thus, the length of the AP

= 5 \sqrt{10}

units.
Hence, option 1 is correct.
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