Question

Write an equation of a circle with center {-1,2} passing through the point {2,4}

Answer 1

Answer:

The equation of circle passing through points  (2, 4) and center (-1 , 2) is (x  + 1 )² + (y - 2)² = 13

Step-by-step explanation:

Given as :

The circle having center = (-1 , 2)

The circle passes through point = (2, 4)

Now, Standard equation of circle with center and passing through points is

(x - h)² + (y -k)² = r²

where h and k are the center of circle and r is the radius of circle

Now the center as (h,k) = (-1 , 2)

And passing through points (x,y) = (2, 4)

Now, satisfying the center and points on standard circle equation

I.e (x - h)² + (y -k)² = r²

Or, (2 - (-1) )²+ (4 -2)² = r²

or, 3² + 2² = r²

or,  r² = 9 + 4

Or ,  r² = 13

∴  r = $\sqrt{13}$

Now circle equation

(x - (-1) )² + (y -2)² = ($\sqrt{13}$)²

or, (x  + 1 )² + (y - 2)² = 13

So, equation of circle is (x  + 1 )² + (y - 2)² = 13

Hence The equation of circle passing through points  (2, 4) and center        (-1 , 2) is (x  + 1 )² + (y - 2)² = 13   Answer

Answer 2

Answer:

The equation of the circle is given by,

$(x-2)^{2} + (y-4)^{2} = (\sqrt{13})^{2}$

Step-by-step explanation:

The center of the circle is given as (-1,2) and it passes through (2,4).

The distance between the two points is radius of the circle.

$(Distance) = \sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

Radius = $\sqrt{(2-(-1))^{2}+(4-2)^{2}}$

Radius = $\sqrt{13}$

The equation of the circle is given by,

$(x-x1)^{2} + (y-y1)^{2} = (Radius)^{2}$

Inserting above values,

$(x-2)^{2} + (y-4)^{2} = (\sqrt{13})^{2}$