SAT Math: x² + y² - 8x + 10y - 59 = 0 Tangent Line at Point P(4, 5)
Complete the square: (x - 4)^2 + (y + 5)^2 = 100. Center is (4, -5), radius is 10. The radius from (4, -5) to P(4, 5) is vertical (x = 4), so the perpendicular tangent line is horizontal: y = 5. The y-intercept is b = 5.
Exact Question Text
The equation of a circle in the $xy$-plane is given below:
> $$ x^2 + y^2 - 8x + 10y - 59 = 0 $$
A line tangent to this circle at point $P(4, 5)$ intersects the $y$-axis at $(0, b)$. What is the value of $b$?
Step-by-Step Algebraic Solution
- Complete the Square:
$$ (x^2 - 8x + 16) + (y^2 + 10y + 25) = 59 + 16 + 25 $$
$$ (x - 4)^2 + (y + 5)^2 = 100 $$
- Center $(h, k) = (4, -5)$
- Radius $r = \sqrt{100} = 10$
- Find the Radius Line Slope:
From $(4, -5)$ to $P(4, 5)$, $x = 4$ is constant $\implies$ vertical line.
- Determine the Tangent Line:
Perpendicular to a vertical line is a horizontal line: $y = 5$. The line crosses the $y$-axis at $(0, 5) \implies b = \mathbf{5}$.
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