Question

can you help me to find the eqaution,center and radius? : center at origin, circle passes through (4,1)

Answer 1

general equation is x^2 +y^2 = r^2
for center at origin.

Answer 2

here radius r is the distance between (0,0) and (4,1).

Answer 3

First step is to find the radius.
You have to points: (0,0) and (4,1)
\[\sqrt{[(4-0)^{2}+(1-0)^{2}}\]
=\[\sqrt{17}\]

so radius= square root of 17

Next, use the fomular\[(x-a)^{2}+(y-b)^{2}=r ^{2}\]

\[(x-4)^{2}+(y-1)^{2}=17\]

Open up the brackets and...voila!

\[x ^{2}-8x+y ^{2}-2y=0\]

Answer 4

To find the centre:

\[[(x _{1}+x _{2})]\div2, [(y _{1}+y _{2})]\div2\]

\[([0+4]\div2), ([0+1]\div2)\]
(2,0.5) = centre

Answer 5

Thank you for helping me guys!

Answer 6

@Mokeira This is not right for the above question. here it is given that center at origin. So the equation of the circle will be x^2 + y^2 = 17

Answer 7

so the center is (0,0) and raduis is 17? @mikurout

Answer 8

@SeniorHigh yes, center is ( 0,0)
but radius is square root of 17.

Answer 9

@mikurout Oh yeah! I am sorry! my bad...thats correct

Answer 10

centre is (0,0) and radius is \[\sqrt{17}\]

Answer 11

how do you geet the 17?

Answer 12

@SeniorHigh check in my first answer the explanation for radius is correct

Answer 13

Use the distance formula and find the distance between the center (0,0) and a point on the circle (4,1).