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

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

  2. 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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