Question

Prove that the points (0,-5),(4,3),and (-4,-3) lie on the circle centred at the origin with radius 5 ...using tis formula d (x2-x1)2 +(y2-y1)2

Answer 1

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 $d_(_0_,_0_)_,_(_c_,_d_) = \sqrt{(c)^{2} + (d) ^{2}} = 5$

[tex]d_(_0_,_0_)_,_(_0_,_-5_) = \sqrt{(0)^{2} + (-5) ^{2}} = \sqrt{(0) + (25)} = \sqrt{25} = 5 \\ \\ d_(_0_,_0_)_,_(_4_,_3_) = \sqrt{(4)^{2} + (3) ^{2}} = \sqrt{(16) + (9)} = \sqrt{25} = 5 \\ \\ d_(_0_,_0_)_,_(_-4_,_-3_) = \sqrt{(-4)^{2} + (-3) ^{2}} = \sqrt{(16) + (9)} = \sqrt{25} = 5[/tex]

Thus all the three points are on the circle.