Answer to Question #171137 in Calculus for Jon jay mendoza

Question

1. Determine the general form of the equation of the circle whose center is (3,2) and

whose graph contains the point (1,2). Sketch the graph.

2.Write the general equation of the circle whose center is (5, -1) and whose radius is 4.

Sketch the graph.

3.Convert the equation x^2 + y^2 + 2x − 6y + 7 = 0 to standard form and sketch the graph.

4.4. Change the equation (x − 2)^2 + (y − 6)^2 = 9 to general form and sketch the graph.

Answer 1

1) Use distance formula to find the radius of the circle as,

$r=\sqrt{(1-3)^2+(2-2)^2}=\sqrt{2^2}=2$

The standard form of the circle is,

$(x-3)^2+(y-2)^2=2^2$

Expand to obtain the general form of the circle as,

$x^2+3^2-6x+y^2+2^2-4y=4$

$x^2+y^2-6x-4y+9=0$

Therefore, the general form of the circle is $x^2+y^2-6x-4y+9=0$

The sketch of the circle is as shown in the figure below:

2) The standard equation of the circle with center at $(5,-1)$ having radius $4$ is,

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

Expand to obtain the general form of the circle as,

$x^2+5^2-10x+y^2+1^2+2y=16$

$x^2+y^2-10x+2y+10=0$

Therefore, the general equation of the circle is $x^2+y^2-10x+2y+10=0$

The sketch of the circle is as shown in the figure below:

3) Consider the general form of the circle $x^2+y^2+2x-6y+7=0$

Rewrite the circle in standard form as,

$(x^2+2(1)(x)+1^2-1^2)+(y^2-2(3)(y)+3^2-3^2)+7=0$

$(x+1)^2+(y-3)^2-1-9+7=0$

$(x+1)^2+(y-3)^2=3$

Therefore, the standard form of the circle is $(x+1)^2+(y-3)^2=3$

The sketch of the circle is as shown in the figure below:

4) Consider the standard form of the circle $(x-2)^2+(y-6)^2=9$

Expand to obtain the general equation of the circle as,

$x^2+2^2-4x+y^2+36-12y=9$

$x^2+y^2-4x-12y+31=0$

Therefore, the general form of the circle is $x^2+y^2-4x-12y+31=0$

The sketch of the circle is as shown in the figure below: