Conics Section: Equation of a circle

Question

I have a question that states

Find the equation of a circle with center $(2,-3)$ and passing through $(3, -5)$.

I arrived at $x^2 + y^2 -4x + 6y - 5 = 0 $. It was marked wrong. The answer is said to be $x^2 + y^2 -4x + 6y +8 = 0$.

My question is isn't the approach of solving this question same as $(x -a)^2 + (y-b)^2 = (x₁-a)^2 + (y₁-b)^2$ where $(a,b)$ is the coordinate of the center of the circle and $(x_1, y_1)$ is the coordinate of the point.

Answer 1

The circle has the equation

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

where $r$ is unknown. The point $(3,-5)$ is on the circle, hence

$$(3-2)^2+(-5+3)^2=r^2.$$

This gives $r^2=5.$

Hence

$$(x-2)^2+(y+3)^2=5$$

or

$$x^2 + y^2 -4x +6y +8 =0.$$

Answer 2

For a quick check,

$$3^2+(-5)^2-4\cdot3+6(-5)-5\ne 0$$

and

$$3^2+(-5)^2-4\cdot3+6(-5)+8=0.$$

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The constant term in the equation must be

$$a^2+b^2-(x_1-a)^2-(y_1-b)^2=4+9-1-4=8.$$

As noticed by others,

$$-1-4=-5.$$

Answer 3

Yes, you can use the formula $$ (x-a)^2+(y-b)^2 = (x_1-a)^2+(y_1-b)^2 $$ where $(a,b)$ is the centre of the circle and $(x_1,y_1)$ is a point that lies on the circle. This formula works because the RHS tells you the length of the $\text{radius}^2$, which is what you need. Hence, the equation of the circle is $$ (x-2)^2+(y+3)^2=(3-2)^2+(-5+3)^2=5 \, . $$ If we expand the brackets, we get $$ x^2-4x+4+y^2+6y+9=5 $$ which simplifies to the textbook answer.

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Personally, I prefer methods that require less memorisation. Hence, if I was solving this question, I would find the length of the radius using the distance formula, and then write the equation of the circle in the form $$ (x-a)^2 + (y-b)^2 = r^2 \, , $$ before expanding the brackets if necessary.