The centre of a circle is \((2a,\,a-7)\). If it passes through \((11,-9)\) and has diameter \(10\sqrt2\), find \(a\).
\(a=3\) or \(a=5\)
Step 1: The diameter of the circle is given as \(10\sqrt{2}\).
So, the radius is half of the diameter:
\[ r = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2} \]
Step 2: The formula for the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 3: The circle passes through the point \((11, -9)\). So, the distance between the centre \((2a, a - 7)\) and the point \((11, -9)\) must be equal to the radius.
\[ (11 - 2a)^2 + (-9 - (a - 7))^2 = (5\sqrt{2})^2 \]
Step 4: Simplify the right-hand side:
\[ (5\sqrt{2})^2 = 25 \times 2 = 50 \]
Step 5: Expand the terms:
\[ (11 - 2a)^2 + (-9 - a + 7)^2 = 50 \]
\[ (11 - 2a)^2 + (-a - 2)^2 = 50 \]
Step 6: Expand each square:
\[ (11 - 2a)^2 = (2a - 11)^2 = 4a^2 - 44a + 121 \]
\[ (-a - 2)^2 = (a + 2)^2 = a^2 + 4a + 4 \]
Step 7: Add them:
\[ 4a^2 - 44a + 121 + a^2 + 4a + 4 = 50 \]
\[ 5a^2 - 40a + 125 = 50 \]
Step 8: Simplify:
\[ 5a^2 - 40a + 75 = 0 \]
Step 9: Divide the whole equation by 5:
\[ a^2 - 8a + 15 = 0 \]
Step 10: Factorize:
\[ (a - 3)(a - 5) = 0 \]
Step 11: Solve for \(a\):
\[ a = 3 \quad \text{or} \quad a = 5 \]