SAT Math: x² + y² - 8x + 10y - 59 = 0 Tangent Line at Point P(4, 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

1. 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 $r = \sqrt{100} = 10$

1. Find the Radius Line Slope:

From $(4, -5)$ to $P(4, 5)$, $x = 4$ is constant $\implies$ vertical line.

1. Determine the Tangent Line:

Perpendicular to a vertical line is a horizontal line: $y = 5$. The line crosses the $y$-axis at $(0, 5) \implies b = \mathbf{5}$.