What is the equation of a circle with center (6,4) that passes through the coordinate (2,1)?

Question

campbell_st · Answer

well the standard form of the circle is 

$$(x - h)^2 + (y - k)^2 = r^2$$

where (h, k) is the centre and r is the radius. 

you have been given the centre (6, 4)  so substitute those values

then to find the radius.. substitute the point on the curve

(2, 1)

hope it helps

anonymous · Answer

thannk you! could you please help me Rewrite 2x^2+2y^2-8x+10y+2=0 in standard form. Find the center and radius of the circle.  @Campbell_st

campbell_st · Answer

divide every term by 2 and you get 

$$x^2 + y^2 - 4x + 5y + 1 = 0$$

now group them 

$$(x^2 - 4x) + (y^2 + 5y) = -1$$

complete the square in both x and y

then factor for the centre... 

when completing the square, add the same value to both sides of an equation.