The line 2x  y + 6 = 0 meets the circle x^2 + y^2  2y  9 = 0 at points A and B. Find the equation of the circle with AB as diameter.
Question
The line $2x  y + 6 = 0$ meets the circle $x^{2} + y^{2}  2y  9 = 0$ at points $A$ and $B$. Find the equation of the circle with $AB$ as diameter.
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Answer
To find the equation of the circle with diameter $AB$, given the line and circle equations:
Given Equations

Line: $$ 2x  y + 6 = 0 \quad \ldots \text{(i)} $$

Circle: $$ x^2 + y^2  2y  9 = 0 \quad \ldots \text{(ii)} $$
Finding Points of Intersection
First, we need the points where the line intersects the circle. Substituting the expression for $y$ from equation (i) into equation (ii):
$$ y = 2x + 6 $$
Substitute this into (ii):
$$ x^2 + (2x + 6)^2  2(2x + 6)  9 = 0 $$
Expanding and simplifying:
$$ \begin{aligned} x^2 + 4x^2 + 24x + 36  4x  12  9 &= 0 \ 5x^2 + 20x + 15 &= 0 \end{aligned} $$
Dividing by 5:
$$ x^2 + 4x + 3 = 0 $$
Factorizing:
$$ (x + 3)(x + 1) = 0 $$
Thus, $x = 3$ or $x = 1$. From $y = 2x + 6$, we get the corresponding $y$ values:

For $x = 3$: $$ y = 2(3) + 6 = 0 $$ So, point $A$ is $(3, 0)$.

For $x = 1$: $$ y = 2(1) + 6 = 4 $$ So, point $B$ is $(1, 4)$.
Equation of the Circle with $AB$ as Diameter
Given $A(3, 0)$ and $B(1, 4)$, they form the diameter of the new circle.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$ (x  x_1)(x  x_2) + (y  y_1)(y  y_2) = 0 $$
Using our points:
$$ (x + 3)(x + 1) + (y  0)(y  4) = 0 $$
Expanding, we get:
$$ x^2 + 4x + 3 + y^2  4y = 0 $$
Thus, the equation of the circle with $AB$ as diameter is:
$$ \boxed{x^2 + y^2 + 4x  4y + 3 = 0} $$
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