Relation between chords length and radius of circle Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly

Let $P$ be the point where the two chords (and a diameter) meet. Let $h$ (and $k$) be the distance from $P$ to the midpoint of the $2a$ chord (respectively, the $2b$ chord); that is, say $P$ divides the chord into sub-segments of length $a+h$ and $a-h$ (respectively, $b+k$ and $b-k$). Note that $P$ divides a diameter into sub-segments of length $r+c$ and $r-c$ (where $r$ is the radius of the circle); note also that $c$ is the hypotenuse of a right triangle with legs $h$ and $k$: so, $c^2 = h^2 + k^2$.
The Power of a Point principle says that every chord through a particular point of a circle is divided into sub-segments such that the product of the lengths of those sub-segments is a constant (the so-called "power" of the point in question). Thus,
$$(a+h)(a-h) = (b+k)(b-k) = (r+c)(r-c)$$
More succinctly,
$$a^2 - h^2 \;\;=\;\; b^2 - k^2 \;\;=\;\; r^2 - c^2$$
With an eye towards combining an $h^2$ with a $k^2$, I'll add the left-hand and "middle-hand" sides together; their sum is necessarily twice the right-hand side:
$$\begin{align}
( a^2 - h^2 ) + ( b^2 - k^2 ) &= 2 (r^2 - c^2) \\ 
\implies a^2 + b^2 - ( h^2 + k^2 ) &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 - c^2 &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 + c^2 &= 2 r^2
\end{align}$$
