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  • The distance of the mid point of the line joining two points (4,0) and (0,4) from the centre of the circle   <img alt="\mathrm{x^2+y^2=16}" src="https://entrancecorner.oncodecog

The distance of the mid point of the line joining two points (4,0) and (0,4) from the centre of the circle   \mathrm{x^2+y^2=16}   is

Option: 1

\sqrt{2}


Option: 2

2 \sqrt{2}


Option: 3

3 \sqrt{2}


Option: 4

2 \sqrt{3}


Answers (1)

best_answer

Midpoint of the line joining (4, 0) and (0, 4) is

\quad\left(\frac{0+4}{2}, \frac{4+0}{2}\right) \\
or A(2,2)

Centre of circle, \mathrm{x^2+y^2=16}  is   B(0,0)

\mathrm{\therefore \quad \text { Required distance } =A B=\sqrt{(2-0)^2+(2-0)^2} }
\mathrm{ =\sqrt{4+4}=2 \sqrt{2} }
 

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Gunjita

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