Answers

2014-08-26T01:09:57+05:30
As center of circle is O(0, 0) and radius (R) is 5. Now consider any point P(x, y), we have studied following three conditions
(1) OP < R, point P is inside the circle
(2) OP = R, point P is on the circle
(3) OP > R, point P is outside the circle

Now you need to prove that OP = 5.

Formula :
d_(_a_,_b_)_,_(_c_,_d_) =  \sqrt{(a - b)^{2} + (c - d) ^{2}}\\  \\d_(_0_,_0_)_,_(_c_,_d_) = \sqrt{(c)^{2} + (d) ^{2}}

So you need to prove d_(_0_,_0_)_,_(_c_,_d_) = \sqrt{(c)^{2} + (d) ^{2}} = 5

d_(_0_,_0_)_,_(_0_,_-5_) = \sqrt{(0)^{2} + (-5) ^{2}} = \sqrt{(0) + (25)} = \sqrt{25} = 5 \\  \\ &#10;d_(_0_,_0_)_,_(_4_,_3_) = \sqrt{(4)^{2} + (3) ^{2}} = \sqrt{(16) + (9)} = \sqrt{25} = 5 \\  \\ &#10;d_(_0_,_0_)_,_(_-4_,_-3_) = \sqrt{(-4)^{2} + (-3) ^{2}} = \sqrt{(16) + (9)} = \sqrt{25} = 5

Thus all the three points are on the circle.
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