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Step-by-Step Solution
Step 1: Express the circle and identify the intercept lengths
The circle is given by the equation
$ x^2 + y^2 + a x + 2a y + c = 0 \quad (a < 0). $
We know that its intercept on the x-axis is $ 2\sqrt{2} $ and on the y-axis is $ 2\sqrt{5}. $
Step 2: Use the formula for intercepts on the axes
• Intercept on the x-axis occurs where $ y = 0 $. The equation reduces to
$ x^2 + a x + c = 0. $
• The length of the intercept on the x-axis is $ 2 \sqrt{ \frac{a^2}{4} - c } $, which is given to be $ 2\sqrt{2}. $
Thus,
$ 2 \sqrt{ \frac{a^2}{4} - c } = 2\sqrt{2} \ \Longrightarrow\ \sqrt{ \frac{a^2}{4} - c } = \sqrt{2}. $
Squaring both sides leads to
$ \frac{a^2}{4} - c = 2 \ \Longrightarrow\ a^2 - 4c = 8. \quad (1) $
• Intercept on the y-axis occurs where $ x = 0 $. The equation reduces to
$ y^2 + 2a y + c = 0. $
• The length of the intercept on the y-axis is $ 2 \sqrt{ a^2 - c } $, which is given to be $ 2\sqrt{5}. $
Thus,
$ 2 \sqrt{ a^2 - c } = 2\sqrt{5} \ \Longrightarrow\ \sqrt{ a^2 - c } = \sqrt{5}. $
Squaring both sides gives
$ a^2 - c = 5. \quad (2) $
Step 3: Solve for a and c
Subtract equation (1) from (2):
From (1): $ a^2 - 4c = 8 $
From (2): $ a^2 - c = 5 $
Subtracting (1) from (2):
$ (a^2 - c) - (a^2 - 4c) = 5 - 8 \ \Longrightarrow\ -c - (-4c) = -3 \ \Longrightarrow\ 3c = -3a. $
It should be recognized that a small adjustment is needed:
Actually, observe that from the original solution steps:
$ (2) - (1) \ \Longrightarrow\ (a^2 - c) - (a^2 - 4c) = 5 - 8, $
which gives
$ -c + 4c = -3, $
or
$ 3c = -3. $
Thus,
$ c = -1. $
Next, substitute $ c = -1 $ into (2):
$ a^2 - (-1) = 5 \ \Longrightarrow\ a^2 + 1 = 5 \ \Longrightarrow\ a^2 = 4. $
Since $ a < 0, $ we get
$ a = -2. $
Step 4: Write the equation of the circle
Substitute $ a = -2 $ and $ c = -1 $ into the circle equation:
$ x^2 + y^2 + a x + 2a y + c = 0 \ \Longrightarrow\ x^2 + y^2 - 2x + 2(-2)y - 1 = 0. $
Therefore, the circle is
$ x^2 + y^2 - 2x - 4y - 1 = 0. $
Step 5: Identify the tangent perpendicular to the line x + 2y = 0
The line $ x + 2y = 0 $ has slope $ -\frac{1}{2}. $
A line perpendicular to it will have slope $ 2. $
Hence, the general form of a tangent line with slope $ 2 $ is
$ 2x - y + \lambda = 0. $
Step 6: Use the condition that the distance from the center to this tangent is the radius
Rewrite the circle to identify its center and radius:
$ x^2 - 2x + y^2 - 4y - 1 = 0. $
Complete the squares:
$ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 1 + 4 + 1 \ \Longrightarrow\ (x - 1)^2 + (y - 2)^2 = 6. $
Thus, the center is $ (1, 2) $ and the radius is $ \sqrt{6}. $
The perpendicular distance from the center $ (1, 2) $ to the line $ 2x - y + \lambda = 0 $ must be $ \sqrt{6}. $
The distance from a point $ (x_0, y_0) $ to a line $ Ax + By + C = 0 $ is
$ \frac{ | A x_0 + B y_0 + C | }{ \sqrt{ A^2 + B^2 } }. $
Here, $ A = 2, B = -1, C = \lambda, x_0 = 1, y_0 = 2. $
Thus,
$ \frac{ | 2(1) - 1(2) + \lambda | }{ \sqrt{ 2^2 + (-1)^2 } } = \sqrt{6}. $
That is,
$ \frac{ | 2 - 2 + \lambda | }{ \sqrt{5} } = \sqrt{6}
\ \Longrightarrow\ \frac{ | \lambda | }{ \sqrt{5} } = \sqrt{6}
\ \Longrightarrow\ | \lambda | = \sqrt{30}. $
Hence, $ \lambda = \pm \sqrt{30}. $
Step 7: Find the shortest distance from the origin to this tangent
The lines of tangency are
$ 2x - y \pm \sqrt{30} = 0. $
The perpendicular distance of the origin $(0,0)$ to $ 2x - y + \lambda = 0 $ is
$ \frac{ | \lambda | }{ \sqrt{ 2^2 + (-1)^2 } } = \frac{ \sqrt{30} }{ \sqrt{5} } = \sqrt{6}. $
This is the required shortest distance.
Final Answer
$ \sqrt{6} $
