Question

Two circles, called c1 and c2, are graphed below. The center of c1 is at the origin, and the center of c2 is the point in the first quadrant where the line y=x intersects c1. Suppose c1 has radius 2. C2 touches the x and y axes each in one point. What are the equations of the two circles?

Answer 1

The equation of
`C_(1)` is:
`x^2+y^2=4`

and the equation of
`C_(2)` is:
`x^2+y^2-2√(2)x-2√(2)y+2=0`

Explanation:

The standard form of circle equation:
`(x-h)^2+(y-k)^2 = r^2` , where
`(h,k)` is the center and
`r` is the radius of the circle.

The center of the circle
`C_(1)` is at origin or at (0, 0) and has radius of 2.

So, the equation of the circle
`C_(1)` will be........

`(x-0)^2+(y-0)^2=(2)^2\\ \\ x^2+y^2=4 ..........................................(1)`

Now, the center of
`C_(2)` is the point in the first quadrant, where the line
`y=x` intersects
`C_(1)`

Solving the equations
`x^2+y^2=4` and
`y=x` , we will get......

`x^2+x^2= 4` (Substituting
`y` as
`x`)

`2x^2= 4\\ \\ x^2=2\\ \\ x=\pm √(2)`

So,
`y=x = \pm √(2)`

Thus, the center of the circle
`C_(2)` will be at
`(√(2),√(2))` (Negative value is ignored as the point is in first quadrant)

Now, the distance between the center of
`C_(2)` and the x-axis is
`√(2)` and the circle touches the x-axis only at one point.

That means, the radius of the circle
`C_(2)` will be also
`√(2)`

So, the equation of the circle
`C_(2)` will be......

`(x-√(2))^2+(y-√(2))^2= (√(2))^2\\ \\ x^2-2√(2)x+2+y^2-2√(2)y+2=2\\ \\ x^2+y^2-2√(2)x-2√(2)y+2=0`