The circle centered at (2, -1) and with radius 4 intersects the circle centered at (2, 5) and with radius sqrt10 at two points A and B. Find (AB)^2

Guest Apr 12, 2018

#1**+2 **

The circle centered at (2, -1) and with radius 4 intersects the circle centered at (2, 5) and with radius sqrt10 at two points A and B. Find (AB)^2

These 2 circles are both centred on the line x=2

The centres are 6 units apart.

formulas

\((x-2)^2+(y+1)^2=16\qquad (1)\\ (x-2)^2+(y-5)^2=10\qquad (2)\\ subtract\\ (y+1)^2-(y-5)^2=6\\ y^2+2y+1-y^2+10y-25=6\\ 12y-24=6\\ 12y=30\\ 2y=5\\ y=2.5 \)

now draw a skech becasue I am going to look at the top triangle.

\((MB)^2=10-2.5^2=3.75\\ MB=\sqrt{3.75}=\sqrt{\frac{15}{4}}=\frac{\sqrt{15}}{2}\\ AB=2*\frac{\sqrt{15}}{2}\\ AB=\sqrt{15}\\ AB^2=15\)

there may well have been a much simpler fully algebraic way to do this :)

Melody
Apr 12, 2018