Question

A circle that is centered at the origin contains the point (0,4). How can you prove or disprove that the point (2,6) also lies on the circle? Does the point (2,6) lie on the circle?

Answer 1

Since the centre is at the origin, (h, k) = (0, 0)
Hence, we can substitute h and k as 0 respectively.

General form: $(x - 0)^{2} + (y - 0)^{2} = r^{2}$
Now, we know that (0, 4) lies on the circle. Thus, the radius must be 4 units.

So, our circle becomes:
$x^{2} + y^{2} = 16$

Now, substitute (2, 6) to see if it lies on the circle.
$LHS = (2)^{2} + (6)^{2} = 4 + 36 = 40 \neq 16$
Hence, it doesn't lie on the circle.

Answer 2

Answer:

The point  $(2,6)$ does not lie on the circle

Step-by-step explanation:

Step 1

Find the equation of the circle

we know that

The equation of the circle in center radius form is equal to

$(x-h)^{2}+(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have that the center is at the origin so

$(h,k)=(0,0)$

substitute

$(x-0)^{2}+(y-0)^{2}=r^{2}$

$(x)^{2}+(y)^{2}=r^{2}$

The point $(0,4)$ lie on the circle

so

substitute in the equation and solve for r

$(0)^{2}+(4)^{2}=r^{2}$

$(4)^{2}=r^{2}$ --------> $r=4\ units$

The equation of the circle is equal to

$(x)^{2}+(y)^{2}=16$

Step 2

Verify if the point $(2,6)$ lie on the circle

we know that

If a ordered pair lie on the circle, then the ordered pair must be satisfy the equation of the circle

Substitute the value of x and the value of y in the equation and then compare

$(2)^{2}+(6)^{2}=16$

$40=16$ ------> is not true

therefore

The point  $(2,6)$ does not lie on the circle