Location of the circle in relation to a circle:
Let S1 ≡ x2 + y2 + 2g1x + 2f1y+ c1 = 0 and S2 º x2 + y2 + 2g2x + 2f2y+ c1 = 0 be two circles. Let D be discriminant for the quadratic equation in x (or y) which is obtained by eliminating y (or x) from the 2 equations of the circle. Then
(i) they are 2 intersecting circles if D > 0
(ii) they are nonintersecting (having no common points) if D < 0
(iii) they touch each other if D = 0
(iv) If D < 0, that is, the circles are nonintersecting then
(a) S1 = 0 is outside S2 = 0 if S2 (-g1, -f1) > 0 or S1 (-g2, -f2) > 0; equivalently, AB > r1 + r2 here A, B are centres and r1, r2 are radii respectively.
(b) S1 = 0 is inside S2 = 0 if S2 (-g1, -f1) < 0; equivalently, AB < |r2 - r1|
(v) If D = 0, then the circles touches each other
(c) externally if AB = r1 + r2
(d) internally if AB = | r1 - r2|
Chord of contact
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From the point P(x1, y1) outside the circle 2 tangents PA and PB can be drawn to circle. The chord AB joining the points of contact A and B of the tangents from P is called as chord of contact of P(x1, y1) with respect to circle. Its equation can be given by T = 0.
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Illustration: If chord of contact of the tangents drawn to x2+y2=a2 from any point on x2+y2=b2, touches the circle x2+y2=c2, then show that a2=bc
Solution: Assume that P(x1, y1) be any point on x2+y2=b2 that is x12+y12=b2. Equation of the corresponding chord of contact is xx1+yy1-a2=0. It 
Solution:
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