### Video Transcript

Suppose that line segment π΄π΅ is a
diameter of a circle center negative seven, four. If π΅ equals negative eight, six,
what is the general equation of the tangent to the circle at π΄?

Letβs sketch what we know so
far. We have two points of this
circle. Its center is located at negative
seven, four. And point π΅ lies on its outside
edge at negative eight, six. Hereβs point π΅ and the circle
should look something like this. We also know that line segment π΄π΅
is a diameter. And that means line segment π΄π΅ is
created by drawing a line from point π΅ through the center of the circle to point
π΄.

We are looking for the tangent to
the circle at point π΄. The tangent line to point π΄ will
be perpendicular to line segment π΄π΅. To find the tangent, we need a few
things. We need to know the slope of line
π΄π΅ and we need to know where point π΄ is located at.

Letβs start here with the slope of
π΄π΅. To find the slope of any line,
weβll need two points on that line. We can use points π΅ and πΆ:
negative eight, six; negative seven, four. The slope of this line equals π¦
two minus π¦ one over π₯ two minus π₯ 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 π΄π΅ is negative
two over one.

We can use that information to find
point π΄. Point π΄ is negative two down two
and right one from point πΆ, down two, right one. Down two is the π¦-coordinate. Two units down from four is two and
one unit to the right of negative seven is negative six. Point π΄ is located at negative
six, two.

We now have the slope of line π΄π΅
and point π΄. The tangent line to point π΄ must
have a negative reciprocal slope with the slope of line π΄π΅. The negative reciprocal of negative
two is positive one over two. Line segment π΄π΅ and the tangent
at point π΄ are perpendicular lines.

We now have a point and a slope for
our tangent line. So we can use our point-slope
formula: π¦ minus π¦ one equals π times π₯ minus π₯ one. π¦ minus two equals one-half times
π₯ minus negative six. π₯ minus negative six can be
simplified to say π₯ plus six. Bring everything else down. Now, weβre going to distribute our
one-half. One-half times π₯ equals one-half
π₯. One-half times six equals
three. π¦ minus two equals one-half π₯
plus three. We add two to both sides of the
equation. π¦ equals one-half π₯ plus
five.

We have one final step to do. The line π¦ equals one-half π₯ plus
five is the tangent line to point π΄. But itβs not written in the general
format. Itβs not the general equation. So we need to rearrange this to the
general equation format. The general equation format is ππ₯
plus ππ¦ plus π equals zero, where π is a positive and all of the constants are
integers.

Iβll give us a little bit more
space over here. We subtract one-half π₯ from both
sides. Now, we have negative one-half π₯
plus π¦ equals five. Now, weβre going to subtract five
from both sides. Five minus five equals zero. And our left side now reads
negative one-half π₯ plus π¦ minus five.

Remember I want the constant by π₯
to be a positive integer. So Iβm gonna multiply the whole
equation by negative two. Negative two times negative
one-half π₯ equals positive π₯. Negative two times π¦ equals
negative two π¦. 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 π΄ is π₯ minus two π¦ plus 10 equals zero.