### Video Transcript

π· is a circle with equation minus
π₯ squared minus π¦ squared equals minus nine. π is a point on the circle with
coordinates minus 12 over five, nine over five. Find an equation for the line that
passes through the point π and is tangent to the circle π·.

The first thing to notice here is
actually weβre dealing with a circle. So this question is all about the
equation of a circle. Well, the first thing we know about
the equation of a circle is if we have in the general form π₯ squared plus π¦
squared equals π squared, then the center of the circle if itβs in this form is
gonna be equal to zero, zero and the radius of the circle is equal to π. Okay, great, so weβve got this now
and weβll actually use this to help us solve the problem.

So first of all, weβre gonna look
at the equation of our circle, which is minus π₯ squared minus π¦ squared equals
minus nine. Now to actually get it in the form
that weβre looking for so we can actually use our equation of the circle, so the
general form, what weβre gonna do is actually multiply each side of our equation by
minus one. So when we do that, weβre gonna get
π₯ squared plus π¦ squared is equal to nine. And we get that because if you have
minus one multiplied by minus π₯ squared, weβve a minus multiplied by a minus which
gives us a plus, which therefore weβre gonna get plus π₯ squared and the same again
if you multiply minus one by minus π¦ squared and by minus nine.

Okay, great, so weβve now got our
equation and actually itβs in the form that we looked at just now. And actually, what we can do is
actually take one more step and write it this way. So weβve got π₯ squared plus π¦
squared equals three squared. And thatβs because if we look back
to our general form for the equation of a circle, weβve got π₯ squared plus π¦
squared equals π squared. And we know that three squared is
equal to nine.

So therefore, actually what we can
do is use this information now to tell us that our circle is gonna have a center
zero, zero because we said that if itβs in the form π₯ squared plus π¦ squared
equals π squared, then the center is gonna be zero, zero. And also we know the radius and we
know that this is gonna be equal to three. And this is because we said that
actually π₯ squared plus π¦ squared equals π squared, where π being the
radius. Fab, so weβve now got that
information, letβs move on to the next part of the question.

So what Iβve done now just gonna
help us work out what to do in the second part of the question is actually drawing a
little sketch. So weβve got our circle and weβve
got the point π. And what weβre actually looking for
is we actually want to find an equation for the line that passes through the point
π and is a tangent to the circle at π·. So Iβve drawn that line there. So itβs actually gonna be a tangent
to the circle. So what we want to do now is
actually find the gradient of our radius ππ. And we want to do that because
actually itβs gonna be perpendicular to our tangent. So therefore, itβs gonna allow us
to actually work out the gradient of our tangent.

Well, to help us actually find the
gradient of ππ, what we have is this little formula which tells us that π, so our
gradient, is equal to π¦ two minus π¦ one, which is the change in π¦, divided by π₯
two minus π₯ one, which is the change in π₯. Well, we know that weβve actually
got the point here which is gonna be π₯ two, π¦ two which is gonna be minus 12 over
five and nine over five. And our π₯ one, π¦ one are actually
gonna be the center of the circle, so zero, zero.

Okay, so now, letβs substitute
these into our formula to find the gradient. So this is gonna give us the
gradient is equal to nine over five minus zero over minus 12 over five minus zero
which is gonna be equal to nine over five multiplied by five over minus 12. And thatβs because if we divide by
a fraction, thatβs the same as multiplying by the reciprocal. So weβre dividing by minus 12 over
five is the same as multiplying by five over minus 12. So if we divide the top and bottom,
the numerator and the denominator, by five, they cancel out. So weβre left with minus nine over
12, which we can then simplify to give us our gradient π one which is equal to
minus three over four or minus three-quarters.

So now, the next step is to
actually use this to help us find the gradient of the tangent touching π. Okay, so now to find the gradient
of tangent to π· touching π, we can actually use this relationship which is that
the gradient of a line multiplied by the gradient of the line perpendicular to it is
equal to minus one. So in this case, π one multiplied
by π two is gonna be equal to minus one. So therefore, we can say that minus
three over four or minus three-quarters multiplied by π two is equal to minus
one. So therefore, π two is equal to
minus one divided by minus three over four which is equal to four over three. Another way of actually thinking
about this is that if you have the gradient of a line, it is actually equal to the
negative reciprocal of the line thatβs perpendicular to it.

Okay, fab, so now letβs move on to
the next stage. So the next stage is to actually
find the equation of this line, so equation for the tangent of the circle at point
π. Because itβs a straight line, we
know itβs gonna be in the form π¦ equals ππ₯ plus π. And in this case, weβve actually
put π¦ equals π two π₯ plus π because we already know the gradient π two. So we now know that itβs π¦ is
equal to four-thirds of π₯ plus π. And as weβve already said that
because we know that four-thirds is equal to π two.

So now what we need to do is well,
we know that weβve got π¦ equals ππ₯ plus π and π is our gradient, π is in fact
our π¦-intercept. So this is now what we need to
find. And in order to find that, what
weβre actually gonna use are the π₯- and π¦-coordinates of the point that we have
which is point π cause weβre gonna substitute them in for π₯ and π¦. So weβre gonna get nine over five
is equal to four over three multiplied by minus 12 over five plus π which is gonna
give us nine over five is equal to minus 48 over 15 because weβve four multiplied by
minus 12 gives us minus 48 and three multiplied by five gives us 15 then plus
π.

So then, what weβre actually gonna
do is multiply the left-hand side of our equation by three over three because what
we actually want to do is get a common denominator. And the common denominator is
actually gonna be 15. And whatever we do to the bottom,
so we multiply five by three gives 15, we have to do that to the top, so our
numerator, so nine by three gives us 27. So we get 27 over 15 is equal to
minus 48 over 15 plus π. So then, we actually add 48 over 15
to both sides. So we get π is equal to 27 over 15
plus 48 over 15. So then, as we got common
denominators, all we do is we add the numerators. So we get 27 plus 48 which gives us
75. So we can say that π is equal to
75 over 15, which therefore gives us a final answer of π of five.

So therefore, we can say that the
equation for the line that passes through the point π which is minus 12 over five,
nine over five and is tangent to the circle π· is gonna be π¦ equals four-thirds π₯
plus five. And we got the four-thirds because
that was our gradient and we got the five because that was our π¦-intercept.