Answer to Question #182075 in Geometry for Lawrent

Question

The equation of a circle is 4x^2+4y^2-12x+32y=27. Find the chord of the circle that cuts the x-axis.

Answer 1

$4x^2+4y^2-12x+32y=27 \\ 4x^2-12x+9+4y^2+32y+64=27+9+64 \\ (2x-3)^2+(2y+8)^2=10^2 \\ O(3/2;-4) \ R=10 \\ y=0 \\ (2x-3)^2+(2*0+8)^2=10^2 \\ (2x-3)^2+64=100 \\ (2x-3)^2=36 \\ 2x-3=6 \\ 2x=9 \\x=4.5 \\ 2x-3=-6 \\ 2x=-3 \\ x=-1.5 \\ chord \ AB \ \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1} \\ A(x_1;x_2), \ B(y_1;y_2) \\ x_1\in(-1.5;4.5) \ y_1 \in(0;1) \\ x_2 \in[-3.5;-1.5) \cup(4.5;6.5] \\ y_2\in[-9;0)$