Question

Which equation represents the circle described?

Answer 1

We have the equation:

$x^2+y^2-8x-6y+24=0$

By arranging this equation in terms of x and y, we have:

$x^2-8x+y^2-6y=-24$

By using the method of completing the square, we have:

$x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}$

The center of this circle is:

$(h,k)=(4,3)$

So the equation that fulfills the statement is:

$(x-4)^2+(y-3)^2=2^2$

Finally, the right answer is c)

Answer 2

Answer:

The correct option is c.

Step-by-step explanation:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$          .... (1)

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

The given equation is

$x^2+y^2-8x-6y+24=0$

$(x^2-8x)+(y^2-6y)+24=0$

To make a perfect square add and subtract $(-\frac{b}{2a})^2$ in the parenthesis. where, b is coefficient of middle term and a is leading coefficient.

$(x^2-8x+4^2-4^2)+(y^2-6y+3^2-3^2)+24=0$

$(x^2-8x+4^2)-16+(y^2-6y+3^2)-9+24=0$

$(x-4)^2+(y-3)^2-1=0$

Add 1 both sides.

$(x-4)^2+(y-3)^2=1$        . ... (2)

On comparing (1) and (2), we can say that the center of the circle is (4,3).

The equation of circle having center (4,3) with radius 2 is

$(x-4)^2+(y-3)^2=2^2$

Therefore option c is correct.