# If y = 2x forms a chord in the circle x^2 + y^2 - 10x = 0 , find the equation of a circle with this chord as the diameter.

(i)We are given the circle `x^2+y^2-10x=0` . By completing the square we can rewrite this in standard form:

`x^2-10x+y^2=0`

`x^2-10x+25+y^2=25`

`(x-5)^2+y^2=5^2`

This is a circle centered at (5,0) with radius 5.

(ii) The line y=2x forms a chord of this circle. The intersections of the line and the circle are...

## Check Out This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

(i)We are given the circle `x^2+y^2-10x=0` . By completing the square we can rewrite this in standard form:

`x^2-10x+y^2=0`

`x^2-10x+25+y^2=25`

`(x-5)^2+y^2=5^2`

This is a circle centered at (5,0) with radius 5.

(ii) The line y=2x forms a chord of this circle. The intersections of the line and the circle are the endpoints of the chord. Substituting y=2x into the equation for the circle yields:

`x^2+(2x)^2-10x=0`

`5x^2-10x=0`

`5x(x-2)=0`

Thus `x=0 ==> y=0` or `x=2==>y=4`

Let A be the point (0,0) and B the point (2,4)

(iii) Now `bar(AB)` is the diameter of the circle we seek. The center will be at the midpoint of `bar(AB)` , and the radius will be `1/2AB` .

The midpoint of `bar(AB)` is the point (1,2) and the distance from A to the midpoint (the radius) is found using the Pythagorean theorem to be `sqrt(5)` .

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

---------------------------------------------------------------

Thus the equation of the circle we seek is

`(x-1)^2+(y-2)^2=5`

---------------------------------------------------------------

Approved by eNotes Editorial Team