Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C\(_1\)(\(\alpha\), \(\beta\)) and C\(_2\)(\(\gamma\), \(\delta\)), C\(_1\) \(\neq\) C\(_2\) are their centres, then \(|(\alpha+\beta)(\gamma+\delta)|\) is equal to _________.
Show Hint
A point at a distance 'r' from \((x_0, y_0)\) along a line with slope \(m = \tan\theta\) can be found using parametric coordinates: \(x = x_0 \pm r\cos\theta\) and \(y = y_0 \pm r\sin\theta\). From \(m=3/4\), we have a 3-4-5 triangle, so \(\cos\theta=4/5\) and \(\sin\theta=3/5\). The centers are \((1 \pm 5(4/5), 2 \pm 5(3/5))\), which gives \((1\pm4, 2\pm3)\), leading to (5, 5) and (-3, -1).
Step 1: Understand the geometry of the system.
The two circles touch externally at point P(1, 2). The given line \(4x+3y=10\) is the common tangent at this point P. The line connecting the centers, \(C_1\) and \(C_2\), must pass through the point of tangency P and be perpendicular to the common tangent.
Step 2: Find the equation of the line connecting the centers.
The slope of the common tangent \(4x+3y-10=0\) is \(m_T = -4/3\).
The line \(C_1C_2\) is normal to the tangent, so its slope is \(m_N = -1/m_T = 3/4\).
The line passes through P(1, 2). Its equation is:
\[ y - 2 = \frac{3}{4}(x - 1) \]
Step 3: Find the coordinates of the centers.
The centers \(C_1\) and \(C_2\) are on this normal line, at a distance of the radius (r=5) from the point P(1, 2).
Let a center be C(x, y). We can use parametric form for the line or solve a system of equations.
From the line equation, \(y-2 = \frac{3}{4}(x-1)\). Let \(\frac{x-1}{4} = \frac{y-2}{3} = k\).
Then \(x = 1+4k\) and \(y = 2+3k\).
The distance from P(1, 2) to C(x, y) is 5:
\[ \sqrt{(x-1)^2 + (y-2)^2} = 5 \]
\[ \sqrt{(4k)^2 + (3k)^2} = 5 \]
\[ \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = |5k| = 5 \]
This gives \(k = \pm 1\).
Step 4: Calculate the coordinates for k=1 and k=-1.
- For k = 1:
\(x = 1 + 4(1) = 5\)
\(y = 2 + 3(1) = 5\)
So, \(C_1 = (5, 5)\). Thus, \(\alpha=5, \beta=5\).
- For k = -1:
\(x = 1 + 4(-1) = -3\)
\(y = 2 + 3(-1) = -1\)
So, \(C_2 = (-3, -1)\). Thus, \(\gamma=-3, \delta=-1\).
Step 5: Compute the final expression.
We need to find \(|(\alpha+\beta)(\gamma+\delta)|\).
\[ \alpha+\beta = 5+5 = 10 \]
\[ \gamma+\delta = -3 + (-1) = -4 \]
\[ |(\alpha+\beta)(\gamma+\delta)| = |(10)(-4)| = |-40| = 40 \]
