Find the center,radius, intercepts and graph the circle.

Question

x^2 + y^2 + 12x + 10y + 61 = 16

Answer 1

Given equation x^2+y^2+12x+10y+61 = 16

x^2+12x+y^2+10y+36+25 = 16

x^2+12x+36+y^2+10y+25 = 16

(x+6)^2+(y+5)^2 = 4^2 ---------> (1)

(x-(-6))^2+(y-(-5))^2 = r^2

Compare it to standard equation (x-h)^2+(y-k)^2 = r^2

Center (h,k) = (-6,-5)

Radius is 4.

To find x intercept substitite y = 0 in (1).

(x+6)^2+25 = 4^2

(x+6)^2 = 16-25

(x+6)^2 = -9

x+6 = √-9

x = -6+√-9

Squre root of negitive is not possible so no x inetercept here.

To find y intercept substitite x = 0 in (1).

36+(y+5)^2 = 4^2

(y+5)^2 = 16-36

(y+5)^2 = -20

y+5 = √-20

y = -5+√-20

Squre root of negitive is not possible so no y inetercept here.

Comments

Draw the graph of circle.

Answer 2

Standard form of a circle equation is

Where center is (h ,k ) and radius of circle is r .

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here x coefficient = 12. so, (half the x coefficient)² = (12/2)²= 36

Here y coefficient = 10. so, (half the y coefficient)² = (10/2)²= 25

Add 36 and 25 to each side

----> (1)

Compare it to standard form of circle

To find y  intercept substitute x = 0 in the equation (1).

Negative squre root is imaginary,So there is no y  - intercept.

To find x  intercept substitute y  = 0 in the equation (1).

Negative squre root is imaginary,So there is no x - intercept.

Center at (-6,-5) and r  = 4,So we plot

Up (-6, -5+4) = (-6, -1)

Down (-6, -5-4) = (-6, -9)

Left (-6-4, -5) = (-10, -5)

Right (-6+4, -5) = (-2, -5)

Graph

1. draw the coordinate plane.

2. Plot the center at (-6,-5).

3. Plot 4 points " radius away from the center in the up, down, left and right direction.

4. Sketch the circle.