Where does the circle (x-6)^2+(y-7)^2=4 intersect with y=x+3
Where does the circle (x-6)2+(y-7)2=4 intersect with y=x+3We need to sub y=x+3 into the circle equation giving us an equation in just x:(x-6)2+(x-4)2=4Next we expand out the brackets:x2 -12x+36+x2-8x+16=4Next collect the terms:2x2-20x+48=0Next we need to factorise to solve for x:2(x2-10x+24)=2(x-6)(x-4)=0this gives us x solutions of x=6 and x=4Now we need to sub these back into y=x+3 to get the y coordinates.This gives y=9 and y=7The overall answer:The circle and the line given intersect at the points (6,9) and (4,7)
NL
