Asked by Max
the equation x^2+y^=25 describes a circle with center at the origin and radius 5. The line y=x-1 passes through the circle. Using the substitution method, find the points at which the circle and line intersect.
a) (4,3) and (-4,-3)
b) (3,4) and (-3,-4)
c) (4,-3) and (-3,4)
d) no solution
a) (4,3) and (-4,-3)
b) (3,4) and (-3,-4)
c) (4,-3) and (-3,4)
d) no solution
Answers
Answered by
Steve
well, do the sub:
x^2 + (x-1)^2 = 25
or, even without doing the math, you can see that none of the choices satisfies y=x-1 for both points
But (d) is also wrong, since the line crosses at (-3,-4) and (4,3)
I suspect at least one typo.
x^2 + (x-1)^2 = 25
or, even without doing the math, you can see that none of the choices satisfies y=x-1 for both points
But (d) is also wrong, since the line crosses at (-3,-4) and (4,3)
I suspect at least one typo.
Answered by
Max
Thanks.
That was my problem, I couldn't find a right answer.
Is there a problem with my work?
x^2+y^2=25 y=x-1
x^2+(x-1)^2=25 (x-1)(x-1)=x^2-2x+1
x^2+x^2-2x+1=25
2x^2-2x+1=25
2x^2-2x-24
2(x^2-x-12)
2(x-4)(x+3)
x-4) (x+3)
x=4 x=-3
y=x-1
y=4-1 y=-3-1
y=3 y=-4
(4,3) (-3,-4)
That was my problem, I couldn't find a right answer.
Is there a problem with my work?
x^2+y^2=25 y=x-1
x^2+(x-1)^2=25 (x-1)(x-1)=x^2-2x+1
x^2+x^2-2x+1=25
2x^2-2x+1=25
2x^2-2x-24
2(x^2-x-12)
2(x-4)(x+3)
x-4) (x+3)
x=4 x=-3
y=x-1
y=4-1 y=-3-1
y=3 y=-4
(4,3) (-3,-4)
Answered by
Steve
Looks good to me
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