y=mx is a chord of the circle of radius a thru origin & whose diameter is along the x-axis .?

Question

Find the equ. of the circle whose diameter is the chord.Hence find the locus of its center for all the values of m

Answer 1

I'm not sure how far I'll get with this one, but I'll give it a try.

Let the center of the circle be at (p,0), so the equation of the circle is (x-p)^2 + y^2 = a^2, with a>p. The chord is y = mx, and we want to find where the chord and circle intersect.

(x-p)^2 + m^2 x^2 = a^2
x^2 - 2px + p^2 + m^2 x^2 = a^2
(m^2 + 1) x^2 - 2px = a^2 - p^2

Solve by completing the square (details omitted) to get

x = {p +/- sqrt[m^2 (a^2 - p^2) + a^2]} / (m^2 + 1)

The center of the chord (hence center of the circle with chord as diameter) is at x = p / (m^2 + 1), with y = mx = mp / (m^2 + 1). The radius is sqrt[m^2 (a^2 - p^2) + a^2] / (m^2 + 1), so the equation of that circle is

[x - p / (m^2 + 1)]^2 + [y - mp / (m^2 + 1)]^2
= [m^2 (a^2 - p^2) + a^2] / (m^2 + 1)^2

If q = m^2 + 1, then by clearing fractions, we get

(qx - p)^2 + (qy - mp)^2 = m^2 (a^2 - p^2) + a^2

I'm not going to check my work, but those are the answers I ended up with.

Answer 2

diameter = cord
y=mx,
x^2 + y^2 =a^2
so
x^2 +m^2x^2 =a^2
x^2(1+m^2)=a^2
x^2 = a^2 / (1+m^2)
x= a/sqrt(1+m^2) or x = - a/sqrt(1+m^2)
point on the cord and on the circle are:
(a/sqrt(1+m^2) , am/sqrt(1+m^2) ) and ( - a/sqrt(1+m^2) , -am/sqrt(1+m^2) )
now the diameter of the circle is the distance between these points:
sqrt( (2am/sqrt(1+m^2))^2 + (2a/sqrt(1+m^2))^2)
= sqrt[ ( 4 a^2m^2 + 4a^2 ) / (1+m^2)]
= 2a sqrt[(m^2 +1)/(1+m^2)]
= 2a
therefore the radius is: a
and the center is (0,0) just add the two points and divide by 2.
so the equation of such a circle is the same as the original circle
s

a simplier reason, without so much computation is that
y=mx is not just a cord, it is a diameter of the original circle since it passes through the origin....