Question

Circle with centre (0, and passing
through the projection of (2, 4) on x-axis
is​

Answer 1

The equation of a circle can be represented as :

$(x-a)^2 + (y-b)^2 = r^2 ...... (1)$

Where (x,y) are the co-ordinates from where the circle passes.

(a,b) is the co-ordinate of center of the circle.

r is the radius of circle.

Here, we are given that center of circle is at (0,0).

So, a = 0 and b = 0.

The equation (1) is reduced to:

$\Rightarrow x^2 + y^2 = r^2 ...... (2)$

Also, we are given that the circle passes through the reflection of point (2,4) on x-axis.

The resultant point is (2,-4).

Putting value of x = 2 and y = -4 in equation (2).

$(2)^2 + (-4)^2 = r^2\\\Rightarrow r^2 = 20$

The equation of circle is:  

$x^2 + y^2 = 20$

Please refer to the image attached for the graph of circle.