Question

Find the equation of the sphere which pass through the circle x^2+y^2+z^2=9,2x+2y-7=0& the touch the plane x-y+z+3=3

Answer 1

The sphere

$$x^2+y^2+z^2-9+\lambda(2x+2y-7)=0$$

passes through the given circle for all values of $\lambda.$

Its center is $(-\lambda,\ -\lambda,\ 0 ).$

Radius is $\sqrt{2\lambda^2+7\lambda+9}.$

If a plane $x-y+z+3=0$ touches a sphere then the lenght of the the perpendicular from its centre to the plane is equal to its radius

$${|-\lambda+\lambda+0+3| \over \sqrt{(1)^2+(-1)^2+(1)^2}}=\sqrt{2\lambda^2+7\lambda+9}$$

$$2\lambda^2+7\lambda+9=3$$

$$2\lambda^2+7\lambda+6=0$$

$$\lambda={ -7\pm\sqrt{(7)^2-4(2)(6)}\over 2(2)}={ -7\pm1\over 4}$$ $$\lambda_1=-2,\ \lambda_2=-{3 \over 2}$$

$\lambda_1=-2$

Its center is $(2,\ 2,\ 0 ).$

Radius is $\sqrt{2(-2)^2+7(-2)+9}=\sqrt{3}.$

The equation of the sphere is

$$(x-2)^2+(y-2)^2+z^2=3$$

$\lambda_2=-{3 \over 2}$

Its center is $({3 \over 2},\ {3 \over 2},\ 0).$

Radius is $\sqrt{2(-{3 \over 2})^2+7(-{3 \over 2})+9}=\sqrt{3}.$

The equation of the sphere is

$$(x-{3 \over 2})^2+(y-{3 \over 2})^2+z^2=3$$