The equation of a circle is \(x^2 + y^2 - 4x + 6y - 77 = 0\).
(a) Find the \(x\)-coordinates of the points \(A\) and \(B\) where the circle intersects the \(x\)-axis.
(b) Find the point of intersection of the tangents to the circle at \(A\) and \(B\).
Solution
(a) To find the \(x\)-coordinates where the circle intersects the \(x\)-axis, set \(y = 0\) in the circle's equation:
\(x^2 - 4x - 77 = 0\).
Factor the quadratic equation:
\((x + 7)(x - 11) = 0\).
Thus, the \(x\)-coordinates are \(-7\) and \(11\).
(b) The center of the circle \(C\) is \((2, -3)\). The gradient of \(AC\) is \(\frac{1}{3}\) or the gradient of \(BC\) is \(\frac{1}{3}\).
The gradient of the tangent at \(A\) is \(3\) or the gradient of the tangent at \(B\) is \(-3\).
The equations of the tangents are \(y = 3x + 21\) and \(y = -3x + 33\).
Set the equations equal to find the intersection:
\(3x + 21 = -3x + 33\).
Solve for \(x\):
\(6x = 12\)
\(x = 2\).
Substitute \(x = 2\) into one of the tangent equations to find \(y\):
\(y = 3(2) + 21 = 27\).
Thus, the point of intersection is \((2, 27)\).
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