Video Transcript
Suppose that line segment 𝐴𝐵 is a
diameter of a circle center negative seven, four. If 𝐵 equals negative eight, six,
what is the general equation of the tangent to the circle at 𝐴?
Let’s sketch what we know so
far. We have two points of this
circle. Its center is located at negative
seven, four. And point 𝐵 lies on its outside
edge at negative eight, six. Here’s point 𝐵 and the circle
should look something like this. We also know that line segment 𝐴𝐵
is a diameter. And that means line segment 𝐴𝐵 is
created by drawing a line from point 𝐵 through the center of the circle to point
𝐴.
We are looking for the tangent to
the circle at point 𝐴. The tangent line to point 𝐴 will
be perpendicular to line segment 𝐴𝐵. To find the tangent, we need a few
things. We need to know the slope of line
𝐴𝐵 and we need to know where point 𝐴 is located at.
Let’s start here with the slope of
𝐴𝐵. To find the slope of any line,
we’ll need two points on that line. We can use points 𝐵 and 𝐶:
negative eight, six; negative seven, four. The slope of this line equals 𝑦
two minus 𝑦 one over 𝑥 two minus 𝑥 one, four minus six over negative seven minus
negative eight. Four minus six equals negative
two. Negative seven minus negative eight
is the same thing as negative seven plus eight. And that equals one. The slope of line 𝐴𝐵 is negative
two over one.
We can use that information to find
point 𝐴. Point 𝐴 is negative two down two
and right one from point 𝐶, down two, right one. Down two is the 𝑦-coordinate. Two units down from four is two and
one unit to the right of negative seven is negative six. Point 𝐴 is located at negative
six, two.
We now have the slope of line 𝐴𝐵
and point 𝐴. The tangent line to point 𝐴 must
have a negative reciprocal slope with the slope of line 𝐴𝐵. The negative reciprocal of negative
two is positive one over two. Line segment 𝐴𝐵 and the tangent
at point 𝐴 are perpendicular lines.
We now have a point and a slope for
our tangent line. So we can use our point-slope
formula: 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. 𝑦 minus two equals one-half times
𝑥 minus negative six. 𝑥 minus negative six can be
simplified to say 𝑥 plus six. Bring everything else down. Now, we’re going to distribute our
one-half. One-half times 𝑥 equals one-half
𝑥. One-half times six equals
three. 𝑦 minus two equals one-half 𝑥
plus three. We add two to both sides of the
equation. 𝑦 equals one-half 𝑥 plus
five.
We have one final step to do. The line 𝑦 equals one-half 𝑥 plus
five is the tangent line to point 𝐴. But it’s not written in the general
format. It’s not the general equation. So we need to rearrange this to the
general equation format. The general equation format is 𝑎𝑥
plus 𝑏𝑦 plus 𝑐 equals zero, where 𝑎 is a positive and all of the constants are
integers.
I’ll give us a little bit more
space over here. We subtract one-half 𝑥 from both
sides. Now, we have negative one-half 𝑥
plus 𝑦 equals five. Now, we’re going to subtract five
from both sides. Five minus five equals zero. And our left side now reads
negative one-half 𝑥 plus 𝑦 minus five.
Remember I want the constant by 𝑥
to be a positive integer. So I’m gonna multiply the whole
equation by negative two. Negative two times negative
one-half 𝑥 equals positive 𝑥. Negative two times 𝑦 equals
negative two 𝑦. Negative two times negative five
equals positive 10. Negative two times zero equals
zero.
The general equation of the tangent
to the circle at point 𝐴 is 𝑥 minus two 𝑦 plus 10 equals zero.
