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        The circle whose equation are x^2+y^2+c^2=2ax and x^2+y^2+c^2=2by will touch one another externally,if
2 years ago

							Center is same so both the circles cant touch externally if they are touchimg then they touch infernally

2 years ago
							Here the center are not same of two circle so they cant touch internally they touch externally but I want this condition what. Is the condition for this

2 years ago
							Sorry there is mistake in my answer, we can write circle equation as (x-a)^2 + y^2 = a^2-c^2  so by this  center is (a,0) and radius is √(a^2-c^2) similarly for Nother circle the equation as x^2+(y-b)^2 = b^2-c^2 so the center is (0,b) and radius is √(b^2-c^2) ao if two circles are touching externally then the realtion between them is C1C2=r1+r2 means distance between center = sum of radii of both the circles  so by this √(a^2+b^2)=√(a^2-c^2) + √(b^2-c^2) so further if u solve this u will found out that c^2(a^2+b^2)=a^2.b^2 and it equals to 1/c^2= 1/a^2 + 1/b^2.

2 years ago
							Equation of 1st circle is x2 + y2 – 2ax + c2 = 0.Let C1 be the centre and r1 be the radius of this circle.Clearly, C1  (a,0)  &  r1 = (a2 + 02 – c2) = (a2 – c2)Equation of 2nd circle is x2 + y2 – 2by + c2 = 0.Let C2 be the centre and r2 be the radius of this circle.Clearly, C2  (0,b)  &  r2 = (02 +b2 – c2) = (b2 – c2)The circles touch each other externally. So, the sum of their radii is equal to the distance between their centres,i.e., r1 + r2 = C1C2  => (a2 – c2) + (b2 – c2)  = (a – 0)2 + (0 – b)2Square both sides  =>  (a2 – c2) + (b2 – c2) + 2(a2 – c2)(b2 – c2)  = a2 + b22(a2 – c2)(b2 – c2)  =  2c2Again square both sides  (a2 – c2)(b2 – c2) = c4   or  a2 b2 – a2 c2  – b2 c2 + c4 = c4a2 b2 = a2 c2 + b2 c2              Divide both sides by a2 b2 c2 , to get 1/c2  =  1/a2  +  1/b2

one year ago
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