Q.19. Condition for touching of two given circles.

Question 19.
If the circle (x-a)^2 + (y-b)^2 = c^2 and (x-b)^2 + (y-a)^2 = c^2 touch each other, then prove that 
a=b(+)(-) √c.

Answer 19.
The equations of the given circles are
S(1) : (x-a)^2 + (y-b)^2 = c^2 ... ( 1 ) and
S(2) : (x-b)^2 + (y-a)^2 = c^2 ... ( 2 )

=> equation of the common chord is
S(1) - S(2) = 0
=> (x^2 - 2ax + a^2 + y^2 - 2by + b^2) - (x^2 - 2bx + b^2 + y^2 - 2ay - a^2) = c^2 - c^2
=> 2(b-a)x + 2(a-b)y = 0
=> x - y = 0

If the circles touch each other, the common chord must be the common tangent
Centre of the first circle is (a, b) and radius = c
=> perpendicular distance from the centre of any circle to the tangent = radius of that circle
=> l a - b l / √2 = c
=> (a - b) = ± c√2
=> a = b ± c√2

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