question_answer

Find the radius of the circle which passes through the origin, (0, 4) and (4, 0).

A)  \[2\]                            

B)  \[4\sqrt{2}\]              

C)  \[\sqrt{8}\]          

D)  \[3\sqrt{2}\]

Correct Answer: C

Solution :

(c): Let A (0, 0) B (0, 4) and C (4, 0) be three points on circle with centre \[O({{x}_{1}},{{y}_{1}})\] \[\therefore \] Radius \[=OA=OB=OC\] \[{{({{x}_{1}}-0)}^{2}}+{{({{y}_{1}}-0)}^{2}}={{({{x}_{1}}-4)}^{2}}+{{({{y}_{1}}-0)}^{2}}\] \[\Rightarrow \]\[x_{1}^{2}+y_{1}^{2}-x_{1}^{2}-8{{x}_{1}}+16+y_{1}^{2}\] \[\Rightarrow \]\[{{x}_{1}}=2\] \[{{({{x}_{1}}-0)}^{2}}+{{({{y}_{1}}-0)}^{2}}={{({{x}_{1}}-0)}^{2}}+{{({{y}_{1}}-4)}^{2}}\] \[\Rightarrow \]\[x_{1}^{2}+y_{1}^{2}=x_{1}^{2}+y_{1}^{2}-8{{y}_{1}}+16\] \[\Rightarrow \]\[{{y}_{1}}=2\] \[\therefore \] radius \[=\sqrt{{{(2-4)}^{2}}+{{(2-0)}^{2}}}\] (by distance formula) \[=\sqrt{4+4}=2\sqrt{2}\] unit