Question

9. Consider the circles with the following equations:

Answer 1

Answer with Step-by-step explanation:

We are given that

$x^2+y^2=(\sqrt 2)^2$

$(x-3)^2+(y-3)^2=32=(4\sqrt 2)^2$

Compare with the equation of circle

$(x-h)^2+(y-k)^2=r^2$

Where center of circle=(h,k)

r=Radius of circle

a.Center of circle=(0,0)

Radius=$\sqrt 2$ units

Center of second circle=(3,3)

Radius of second circle=$4\sqrt 2$ units

b.Distance formula:$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Using the formula

The distance between the centers of two circle

=$\sqrt{(3-0)^2+(3-0)^2}=3\sqrt 2$

Hence, the distance between the centers of two circle =$3\sqrt 2$ units.

c.

Substitute x=-1 and -1

$1+1=2=$

$(1-3)^2+(1-3)^2=32$

The circle must be tangent because there is just one point (-1,-1) is common in both circles and satisfied the equations of circle.