Question

A circle with centre at (3, 6) passes through (-1,1). Its equation is

• A. $x^{2}+y^{2}-6x-12y+3=0$
• B. $x^{2}+y^{2}+6x-10y+3=0$
• C. $x^{2}+y^{2}-3x-6y+1=0$
• D. $x^{2}+y^{2}+5x+9y+5=0$
• E. $x^{2}+y^{2}-6x-12y+4=0$

Hint:

Geometry Tip: Expanding $(x-h)^2$ will always yield $-2hx$, and expanding $(y-k)^2$ yields $-2ky$. With center $(3,6)$, the equation must contain $-6x$ and $-12y$. This immediately eliminates options B, C, and D!

Answer 1

Answer: (E)

Concept:
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center coordinate and $r$ is the radius. The radius squared ($r^2$) can be found using the distance formula squared: $r^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ between the center and any point on the circumference.

Step 1: Identify the center coordinates.
The center is given as $(h, k) = (3, 6)$. Substituting this into the standard circle equation gives: $$(x - 3)^2 + (y - 6)^2 = r^2$$

Step 2: Calculate the radius squared.
The circle passes through the point $(x, y) = (-1, 1)$. We find $r^2$ by measuring the squared distance from the center $(3, 6)$ to this point: $$r^2 = (-1 - 3)^2 + (1 - 6)^2$$ $$r^2 = (-4)^2 + (-5)^2$$ $$r^2 = 16 + 25 = 41$$

Step 3: Construct the standard equation.
Substitute $r^2 = 41$ back into our equation from
Step 1: $$(x - 3)^2 + (y - 6)^2 = 41$$

Step 4: Expand the binomial squares.
To match the multiple-choice options, we must expand the equation into its general form. Expand $(x - 3)^2$: $x^2 - 6x + 9$ Expand $(y - 6)^2$: $y^2 - 12y + 36$ Combine them: $$(x^2 - 6x + 9) + (y^2 - 12y + 36) = 41$$

Step 5: Simplify into the general form.
Group the variable terms together and combine all constants on the left side: $$x^2 + y^2 - 6x - 12y + 45 = 41$$ Subtract 41 from both sides: $$x^2 + y^2 - 6x - 12y + 4 = 0$$ Hence the correct answer is (E) $x^{2+y^{2}-6x-12y+4=0$}.