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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

Ajay Pachankar , 7 Years ago

Grade 11

4 Answers

Meet

Last Activity: 7 Years ago

Center is same so both the circles can`t touch externally if they are touchimg then they touch infernally

Ajay Pachankar

Last Activity: 7 Years ago

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

Meet

Last Activity: 7 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.

Samyak Jain

Last Activity: 6 Years ago

Equation of 1^st 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) = $\sqrt{}$ (a²– c²)

Equation of 2^ndcircle 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 = $\sqrt{}$ (0²+b² – c²) = $\sqrt{}$ (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 => $\sqrt{}$ (a²– c²) + $\sqrt{}$ (b²– c²) = $\sqrt{}$ (a – 0)²+ (0 – b)²

Square both sides => (a²– c²) + (b²– c²) + 2 $\sqrt{}$ (a²– c²)(b²– c²) = a²+ b²

2 $\sqrt{}$ (a²– c²)(b²– c²) = 2c²

Again square both sides

$\therefore$ (a²– c²)(b²– c²) = c⁴ or a²b²– a²c² – b²c²+ c⁴ = c⁴

a²b²= a²c²+ b²c² Divide both sides by a²b² c² , to get

1/c² = 1/a² + 1/b²

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