If a circle has center (0,0) and a point on the circle (-2,-4) write the equation of the circle.?

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Answer 1 · 2018-06-20T12:45:48

# x^2+ y^2 =20 #

The general equation of a circle of radius #r# centred on #(a,b)# is:

# (x-a)^2+ (y-b)^2 =r^2 #

Thus a circle centered on the origin will have an equation of the form:

# x^2+ y^2 =r^2 #

Knowing that #(-2,4)# lies on the circle, we have:

# (-2)^2+ (4)^2 =r^2 #

# :. r^2 = 4 + 16 = 20#

Thus the equation is

# x^2+ y^2 =20 #

Answer 2 · 2018-06-20T12:48:41

#x^2+y^2 = 20#

Every point on a circle has the same distance from the center. This distance is the radius #r# of the circle.

So, if #(-2,-4)# is a point on the circle, it means that the radius of the circle is the distance between #(0,0)#, the center, and #(-2,-4)#.

To compute the distance between two points #(x_1y_1)# and #(x_2,y_2)#, the formula is

#d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}#

In this case, #(x_1y_1) = (0,0)# and #(x_2,y_2)=(-2,-4)#. So, their distance is

#d = \sqrt{(0-(-2))^2+(0-(-4))^2} = sqrt(4+16)=sqrt(20)#

Now we know the center #(0,0)# and the radius #sqrt(20)# of the circle. When you have this information, you can write the equation as

#(x-x_0)^2+(y-y_0)^2 = r^2#

where #(x_0,y_0)# is the center and #r# is the radius. So, in this case, the equation is

#(x-0)^2+(y-0)^2 = (sqrt(20))^2#

which can be rewritten as

#x^2+y^2 = 20#