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Grade: 11
        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

Answers : (4)

Meet
137 Points
							Center is same so both the circles can`t touch externally if they are touchimg then they touch infernally
						
2 years ago
Ajay Pachankar
10 Points
							Here the center are not same of two circle so they can`t touch internally they touch externally but I want this condition what. Is the condition for this
						
2 years ago
Meet
137 Points
							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
Samyak Jain
333 Points
							
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 \equiv (a,0)  &  r1 = \sqrt{}(a2 + 02 – c2) = \sqrt{}(a– 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 \equiv (0,b)  &  r2 = \sqrt{}(0+b2 – c2) = \sqrt{}(b– 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  => \sqrt{}(a– c2) + \sqrt{}(b– c2)  = \sqrt{}(a – 0)+ (0 – b)2
Square both sides  =>  (a– c2) + (b– c2) + 2\sqrt{}(a– c2)(b– c2)  = a+ b2
2\sqrt{}(a– c2)(b– c2)  =  2c2
Again square both sides 
\therefore (a– c2)(b– c2) = c  or  ab– a2 c – bc+ c4 = c4
ab= ac+ bc2              Divide both sides by ab2 c2 , to get 
1/c =  1/a +  1/b2
one year ago
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