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.
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briefcaseTeacher (K-12)
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(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.
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Thus the equation of the circle we seek is
`(x-1)^2+(y-2)^2=5`
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