Find the intersection of a vertical line segment in a circle.

Question

My brother needs help coming up with a formula for a problem that I already did but failed to write out the formula for.

The problem is:

Consider a circle with the point (5,4) and a radius of 3. Determine if a vertical line segment with the points (3,5) being one end of the segment and (3,1) being the other end of the segment, intersects this circle. List the intersecting points if so.

: Again I need help coming up with a simple formula to determine where the line intersects the circle.

Thanks

Is $(5,4)$ the center of the circle? (Usually when we say "point" of a circle we mean a point on the circumference, but then you would not have enough information to solve the problem.) — David K, Jan 28 '16 at 03:55
yes 5,4 is the center of the circle. I am looking at the paper right now. I got the right answer I just dont know how. — WhatIsGoingOn, Jan 28 '16 at 03:59

Spencer · Accepted Answer · 2016-01-28T22:57:55.733

1

The points on the line segment each satisfy an equation and two inequalities,

$$ x=3 \qquad y>1 \qquad y<5. $$

The points on the circle each satisfy the equation,

$$ (x-5)^2+(y-4)^2=9. $$

An intersection of these curves will be an ordered pair, $(x,y)$, which solves both equations and satisfies the inequalities.

Since every point on the vertical line segment has the x-coordinate $x=3$ we will start by substituting $x=3$ into the equation of the circle.

$$ (3-5)^2+(y-4)^2=9 $$

$$ 4 + (y-4)^2 = 9 $$ $$ (y-4)^2=5 $$ $$ y-4=\pm\sqrt5 $$ $$ y=4\pm\sqrt5 $$ $$ y \approx 4 \pm 2.23 $$ $$ y\approx 1.77, 6.23$$

This gives us two ordered pairs $(3,1.77)$ and $(3,6.23)$. These both solve the equations for the circle and the line segment, but only one of them satisfies the inequalities. Since $1<y<5$, only the point $(3,1.77)$ is a valid point of intersection.

edited Jan 28 '16 at 22:57

answered Jan 28 '16 at 04:02

Spencer

12,271

Thank you so much but where did you get the 9 from – WhatIsGoingOn Jan 28 '16 at 04:05
The $9$ is the radius of the circle squared. The equation is $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0,y_0)$ is the center and $r$ is the radius. – Spencer Jan 28 '16 at 04:09
I changed the explanation to make it better. If you want to see the old one again you can look at the edit history. Hope this helps :) – Spencer Jan 28 '16 at 04:09
You're amazing thank you so much – WhatIsGoingOn Jan 28 '16 at 04:13

score 1 · Answer 2 · answered Jan 28 '16 at 04:05

The equation for the circle is:

$$ (x - 5)^2 + (y-4)^2 = 9 $$

The equation for the line is:

$$ (3, t) \ \ \ \ t\in[1,5]. $$

This gives:

$$ (3-5)^2 + (t-4)^2 = 4 + t^2 - 8t + 16 = 9 \\ \Rightarrow t^2 -8t + 11 = 0 $$

Using the quadratic formula to solve for $t$ gives:

$$ t = 4 -\pm \sqrt{5}. $$

Since $4 + \sqrt5 > 5$, the only valid value of $t$ is $4 - \sqrt5$, which, after plugging back into the equation for the line, gives the single intersection point:

$$ (3, 4 -\sqrt5) $$

Thank you for your answer it helped a lot! – WhatIsGoingOn Jan 28 '16 at 04:13 — WhatIsGoingOn, Jan 28 '16 at 04:13

Find the intersection of a vertical line segment in a circle.

2 Answers2