Identify the equation of the circle that has its center at (16, 30) and passes through the origin

To solve this question, we have to find the equation of the circle with given center and where it passes. Doing this, we get that the equation of the circle is:

Equation of a circle:

The equation of a circle with center and radius r is given by:

Center at (16, 30)

This means that

Thus

Passes through the origin:

We use this to find the radius squared, as this means that is part of the circle. Thus

Thus, the equation of the circle is:

For another example to find the equation of a circle, you can look at

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

think about it deeply and passionately about this property

i am sorry