Question

Let x, y be real variable satisfy the x²+y²+8x-10y-40 = 0. Let a = max (√(x+2)²+(y-3)²) and b = (√(x+2)²+(y-3)²), then:

• A. a+b=18
• B. a+b=4√2
• C. a-b=4√2
• D. ab=73

Answer 1

Answer: (A), (C), (D)

(A), (C), (D)

Answer 2

1. Convert the circle equation to standard form to find its center $C(-4, 5)$ and radius $r=9$.
2. The expression $\sqrt{(x+2)^2+(y-3)^2}$ is the distance $d$ from a point $P(x,y)$ on the circle to the fixed point $Q(-2,3)$.
3. Calculate the distance $D_{CQ}$ between the circle's center $C$ and the point $Q$: $D_{CQ} = 2\sqrt{2}$.
4. The maximum distance $a$ is $D_{CQ} + r = 2\sqrt{2} + 9$.
5. The minimum distance $b$ is $|D_{CQ} - r| = |2\sqrt{2} - 9| = 9 - 2\sqrt{2}$.
6. Verify the options using the calculated values of $a$ and $b$.