Intersection between a secant/tangent and a circle

[fred2028]

So 2 homework questions I can't get help with anywhere else
1) The line y=x-1 intersects the circle x^2+y^2=25 at 2 points. Find the coordinates of the 2 points.
2) For what value(s) of k will the line y=x+k bea tangent to the circle x^2+y^2=25?

Now my work which I feel like I am off by 5,000,000 miles.

1) 25=x^2+y^2
y^2=25-x^2
y=5-x

The secant is y=x-1

y=x-1 and y=5-x

x-1=5-x
x+x=5+1
2x=6
x=3

This doesn't seem right since there should be 2 results but I only got 1
Now my attempt at question 2 has a similar approach.
2) x^2+y^2=25
y^2=25-x^2
y=5-x

The tangent is y=x+k
x+k=5-x
2x=5-k
Then I get 2 variables with 1 equation ...
God I'm lost

 

[skeeter]

x² + y² = 25
y = x - 1

substitute (x - 1) for y in the circle equation ...

x² + (x - 1)² = 25

now solve for x ... you'll get two solutions.



y = x + k tangent to x² + y² = 25.
one way to do this ...

a tangent line is perpendicular to a circle's diameter ... since the line y = x+k has slope = 1, the diameter must have slope = -1 and pass through the origin.
so ... the line y = -x is the "diameter".

now, find the points where y = -x intersects the circle ...

x² + (-x)² = 25
2x² = 25
x² = 25/2
x = +/- 5/sqrt(2)

points of tangency are (5/sqrt(2), -5/sqrt(2)) and (-5/sqrt(2), 5/sqrt(2))

y = x + k
-5/sqrt(2) = 5/sqrt(2) + k
k = -10/sqrt(2)

y = x + k
5/sqrt(2) = -5/sqrt(2) + k
k = 10/sqrt(2)

 

[fred2028]

substitute (x - 1) for y in the circle equation ...

x² + (x - 1)² = 25

now solve for x ... you'll get two solutions.

For the x^2+(x-1)^2=25, I square rooted all 3 terms, and I got
x+x-1=5
2x-1=5
2x=6
x=3
Which is the same

 

[skeeter]

For the x^2+(x-1)^2=25, I square rooted all 3 terms, and I got
x+x-1=5
2x-1=5
2x=6
x=3
Which is the same

there's your BIG mistake. you do not "square root" them all ...

x² + (x - 1)² = 25

expand (multiply out) the (x - 1)² ...

x² + x² - 2x + 1 = 25

combine like terms ...

2x² - 2x - 24 = 0

divide all terms by 2 ...

x² - x - 12 = 0

factor the quadratic ...

(x - 4)(x + 3) = 0

set each factor equal to 0 and solve ...

x - 4 = 0 ... x = 4
x + 3 = 0 ... x = -3

 

[fred2028]

there's your BIG mistake. you do not "square root" them all ...

O OK thanks!

points of tangency are (5/sqrt(2), -5/sqrt(2)) and (-5/sqrt(2), 5/sqrt(2))

Ummm I don't understand how you got the y values for the tangency points ... I mean I see how you got the x, but not the y ...

 

[Denis]

1) The line y=x-1 intersects the circle x^2+y^2=25 at 2 points. Find the coordinates of the 2 points.
1) 25=x^2+y^2
y^2=25-x^2
y=5-x

Whoa right there: 5-x is NOT the root of 25-x^2:
(5-x)(5-x) = 25 - 10x + x^2 : quite a difference, right!

Since y = x-1, simply substitute that in the circle's equation:
x^2 + (x-1)^2 = 25
x^2 + x^2-2x+1 = 25
2x^2 - 2x - 24 = 0
x^2 - x - 12 = 0
Factor that and you're on your way...

 

[fred2028]

Whoa right there: 5-x is NOT the root of 25-x^2:
(5-x)(5-x) = 25 - 10x + x^2 : quite a difference, right!

Since y = x-1, simply substitute that in the circle's equation:
x^2 + (x-1)^2 = 25
x^2 + x^2-2x+1 = 25
2x^2 - 2x - 24 = 0
x^2 - x - 12 = 0
Factor that and you're on your way...

Lol I get that part, but I don't get how Skeeter got the y values for the coordinates ...

 

[skeeter]

points of tangency are (5/sqrt(2), -5/sqrt(2)) and (-5/sqrt(2), 5/sqrt(2))

Ummm I don't understand how you got the y values for the tangency points ... I mean I see how you got the x, but not the y ...

the points are on the line y = -x , remember?

 

[fred2028]

points of tangency are (5/sqrt(2), -5/sqrt(2)) and (-5/sqrt(2), 5/sqrt(2))

Ummm I don't understand how you got the y values for the tangency points ... I mean I see how you got the x, but not the y ...

the points are on the line y = -x , remember?

Thanks, can't believe I didn't see that. Topic closed ...