Math, asked by behld8227, 4 months ago

The equation of the common tangent to the circle x^2 + y^2 - 4x - 6y - 12 = 0 and
x^2 + y^2 + 6x + 18y + 26 = 0 at their point of contact is

Answers

Answered by Saby123
17

Solution :

The equations of the two circles are :

Circle 1 :

=> x² + y² - 4x - 6y - 12 = 0

Circle 2 :

=> x² + y² + 6x + 18y + 26 = 0 .

Now, first let us calculate the centre and radii of each circle .

Circle 1 :

=> x² + y² - 4x - 6y - 12 = 0

=> x² - 4x + 4 + y² - 6y + 9 - 12 = 4 + 9

=> ( x - 2)² + ( y - 3)² = 25 ....... (1)

Circle 2 :

=> x² + y² + 6x + 18y + 26 = 0 .

=> x² + 6y + 9 + y² + 18y + 81 + 26 = 9 + 81

=> ( x + 3)² + ( y + 9 )² = 73 . ........ (2)

Solving these two equations , we get :

Centre Of First circle => ( 2, 3 ) .

Centre of second circle = ( -3, -9 )

Let the equation of the common tangent of the two circles be T .

=> C₁ C₂ = T²

=> ( x² + y² - 4x - 6y - 12 )( x² + y² + 6x + 18y + 26 ) = T²

T = √ [ ( x² + y² - 4x - 6y - 12 )( x² + y² + 6x + 18y + 26 ) ]

T = √ [ x⁴ + 2x³ + 2 x² y² + 12 x² y - 10x² + 2xy² - 108 xy - 176 x + y⁴ + 12 y³ - 94 y² - 372 y - 312 ]

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