How to show that the locus of centre of circle is a straight line and find its equation?

Interesting problem.

let M (a, b) be the center of ( C1) then the equation of circle C1

is ( x - a)^2 + (y-b)^2 = r ^2 where r is the radius of the (C1)

You try to find the relation between a and b and that is the locus of the center.

Since (C1) pass thru (4, 0 ) then

( 4- a)^2 + b^2 = r^2

Since (C1) and (C2) are orthogonal then

OM^2 = 4 + r^2 ( remember R^2 +r^2 = d^2 formula)

and OM^2 = a^2 + b^2

you arrive at - 8a +16 = -4

or a = 5/2

Since a = 5/2 is a fixed number, then the locus of center of circle (C1) is the line

with equation x = 5/2.

The set of all factors equidistant from the starting place and the point (4,2) is the equation of the perpendicular bisector to the section shaped by utilizing the starting place and the point (4,2). Calculate the segments slope = 2/4 = a million/2 = .5 The midpoint of the section is at (2,a million) and is on the line of the answer The slope of the answer line has a unfavorable reciprocal slope to the section = -2 utilizing the point slope equation, we get y-a million=-2(x-a million) y = -2x + 3