sn0629 發表於 25-8-5 18:41
Circle C₁ has equation x² + y² - 12x - 18y + 76 = 0, circle C₂ has equation x² + y² + 12x + 12 ...
To find the intersection point \((p, q)\) of the two circles \(C_1\) and \(C_2\), given that they intersect at exactly one point, we need to solve their equations simultaneously and ensure the solution yields a single point. The equations are:
\(C_1\): \(x^2 + y^2 - 12x - 18y + 76 = 0\)
\(C_2\): \(x^2 + y^2 + 12x + 12y - 92 = 0\)
Since the circles intersect at exactly one point, they are tangent to each other (either externally or internally). Let’s proceed by finding the intersection point.
Step 1: Rewrite the circle equations in standard form
To understand the geometry, convert the equations to standard form \((x - h)^2 + (y - k)^2 = r^2\) by completing the square.
For \(C_1\): \(x^2 + y^2 - 12x - 18y + 76 = 0\)
Group terms:
\(x^2 - 12x = (x - 6)^2 - 36\)
\(y^2 - 18y = (y - 9)^2 - 81\)
Rewrite:
\[
(x - 6)^2 - 36 + (y - 9)^2 - 81 + 76 = 0
\]
So, \(C_2\) has center \((-6, -6)\) and radius \(r_2 = \sqrt{164} = 2\sqrt{41}\).
Step 2: Check tangency condition
Since the circles intersect at exactly one point, they are tangent. Two circles are tangent if the distance between their centers equals either the sum of their radii (external tangency) or the absolute difference of their radii (internal tangency).
Centers: \(C_1\) at \((6, 9)\), \(C_2\) at \((-6, -6)\).
Distance between centers:
\[
d = \sqrt{(6 - (-6))^2 + (9 - (-6))^2} = \sqrt{(12)^2 + (15)^2} = \sqrt{144 + 225} = \sqrt{369} = 3\sqrt{41}
\]
Radii:
\(r_1 = \sqrt{41}\)
\(r_2 = 2\sqrt{41}\)
Check tangency:
Sum of radii: \(r_1 + r_2 = \sqrt{41} + 2\sqrt{41} = 3\sqrt{41}\)
Difference of radii: \(|r_1 - r_2| = | \sqrt{41} - 2\sqrt{41} | = \sqrt{41}\)
Since the distance \(d = 3\sqrt{41} = r_1 + r_2\), the circles are externally tangent, confirming they intersect at exactly one point.
Step 3: Find the intersection point
To find the point of tangency \((p, q)\), subtract the equations of the circles to eliminate quadratic terms:
\[
(x^2 + y^2 - 12x - 18y + 76) - (x^2 + y^2 + 12x + 12y - 92) = 0
\]
This is the equation of the common chord (or the radical axis). Since the circles are tangent, this line is the tangent line at the point of intersection. Solve this line with one of the circle equations.
Substitute \(y = \frac{28 - 4x}{5}\) into \(C_1\):
\[
x^2 + \left( \frac{28 - 4x}{5} \right)^2 - 12x - 18 \left( \frac{28 - 4x}{5} \right) + 76 = 0
\]
So, the intersection point is \((p, q) = (2, 4)\).
Step 4: Verify the solution
Substitute \((2, 4)\) into both circle equations to confirm.
For \(C_1\):
\[
2^2 + 4^2 - 12(2) - 18(4) + 76 = 4 + 16 - 24 - 72 + 76 = 0
\]
The point \((2, 4)\) satisfies both equations.
To ensure there’s only one intersection, note that the quadratic in \(x\) was a perfect square, yielding a single \(x\)-value, consistent with tangency.
Step 5: Geometric confirmation
The point of tangency lies on the line joining the centers, adjusted by the radii ratio. The line between centers \((6, 9)\) and \((-6, -6)\) has slope:
\[
m = \frac{-6 - 9}{-6 - 6} = \frac{-15}{-12} = \frac{5}{4}
\]
The point \((2, 4)\) should lie on this line or be consistent with the tangency point. Since we’ve solved algebraically and verified, we trust \((2, 4)\).
Final Answer
The intersection point is \((p, q) = (2, 4)\).
\[
\boxed{(2, 4)}
\]
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