At which x values does the circle defined by the equation (x + 2)^2 + (y - 3)^2 = 37 intersect the x-axis when drawn in the xy-coordinate plane?

Question

asked Jan 21, 2024 128k views

2 Answers

Andriy Makukha · Answer 1 · 2024-01-27T07:48:59+0000

Answer:

$x = \left \{ {{{2√(7)-2} } \atop {-2√(7)-2 }} \right.$

= 3.29, x = -7.29

Explanation:

To find an x-intercept, you plug in 0 for y.

(x + 2)^2 + ([0] - 3)^2 = 37

(x+2)^2 + (-3)^2 = 37

(x+2)^2 = 28

Expand:

x^2 + 4x + 4 = 28

x^2 + 4x - 28 = 0

Apply Quadratic Formula: a=1, b=4, c=-24

$x =\left \{ {{\frac{-b+\sqrt{b^(2)-4ac } }{2a} } \atop {\frac{-b-\sqrt{b^(2)-4ac } }{2a} }} \right. \\$

$x =\left \{ {{\frac{-4+\sqrt{4^(2)-4*1*-24 } }{2*1} } \atop{\frac{-4-\sqrt{4^(2)-4*1*-24 } }{2*1} } \right. \\$

$x =\left \{ {{(-4+√(16+96) )/(2)} \atop {{(-4-√(16+96) )/(2)}} \right.$

$x = \left \{ {{(√(112)-4 )/(2) } \atop {(-√(112)-4 )/(2)}} \right.$

$x=\left \{ {{(4√(7)-4 )/(2) } \atop {(-4√(7)-4 )/(2) }} \right.$

$x = \left \{ {{{2√(7)-2} } \atop {-2√(7)-2 }} \right.$

is approximately
$\left \{ {{3.29} \atop {-7.29}} \right.$