Question

Find the equation(s) of the common tangent(s) touching the curves
y² - 6y - 4x + 9 = 0 and x² + y² - 6x – 6y + 9 = 0.​

Answer 1

Step-by-step explanation:

Curve : y² − 6y − 4x + 9 = 0 or

(y - 3)² = 4x or Y² = 4x is a parabola with vertex at (0, 3). An ever tangent to this parabola is given by ; Y = mx + a/m or

(y-3) =mx + (1/m) …. …. … (1) (as a =1) .

If it is a tangent to the curve :

x² + y²−6x −6y + 9 = 0 or

(x - 3)² + (y - 3)²= 9 (a circle), then solving this equation together with eq.(1), we get ; (x-3)² + (mx + 1/m)² = 9 or (1+m²)x² − 4x + 1/m² = 0, a quadratic equation in x and x is real, therefore, for equal roots, it’s discriminant must be 0 . That is, (b² - 4ac) = 16m² − 4(1 + m²) = 0 ==> m = ± (1/√3) . Putting these values of m in eq.(1), we get two equations for the required common tangents as ;

√3y − x = (3 + 3√3) and

√3y + x = (3√3 − 3)