Question

Find the equation of the circle whose radius is 3 and which touches the circle x2+y2-4×-6y-12=0 internally at the point (-1,-1)

Answer 1

Answer:

Step-by-step explanation:

The given circle is

( x - 2 )² + ( y - 3 )² = 12 + 4 + 9 = 25 = 5²

So it has radius 5 and is centred at ( 2, 3 ).

To be tangent, the circles' centres must be collinear with the point of tangency.  So the centre of the new circle is on the line through (-1, 1) and (2, 3).

As the given circle has radius 5 and the new circle has radius 3, the centre is 3/5 of the way from (-1, -1) to (2, 3).  So the centre is at

(3/5) (2, 3) + (2/5) (-1, -1) = ( (6-2)/5, (9-2)/5 ) = ( 4/5, 7/5 ).

Therefore the equation is

( x - 4/5 )² + ( y - 7/5 )² = 3²

Exapnding this to put it in the form of the other circle:

( 5x - 4 )² + ( 5y - 7 )² = 3² × 5²

=>  25x² + 25y² - 40x - 70 y + 16 + 49 - 225 = 0

=>  25x² + 25y² - 40x - 70 y - 160 = 0