question_answer

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 the radius of the circle is

A) \[a+b+c\]

B) \[\frac{\sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{2}\]

C) \[\frac{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}{2}\]

D) \[\frac{\sqrt{ab}}{c}\]

Correct Answer: C

Solution :

[c] From figure In \[\Delta ONA,\]\[O{{N}^{2}}={{r}^{2}}-{{a}^{2}}\]and In \[\Delta ONP,\]\[N{{P}^{2}}=O{{P}^{2}}-O{{N}^{2}}\] \[={{c}^{2}}+{{a}^{2}}-{{r}^{2}}-{{r}^{2}}-{{b}^{2}}\] Now, In \[\Delta OMD,\] \[O{{M}^{2}}=N{{P}^{2}}={{r}^{2}}-{{b}^{2}}\] \[\Rightarrow \]   \[{{c}^{2}}+{{a}^{2}}-{{r}^{2}}={{r}^{2}}-{{b}^{2}}\] \[\Rightarrow \]   \[2{{r}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] \[\Rightarrow \]   \[{{r}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{2}\] \[\Rightarrow \]   \[r=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}{2}\]