Video: Pack 1 β€’ Paper 2 β€’ Question 14

Pack 1 β€’ Paper 2 β€’ Question 14

06:54

Video Transcript

𝐷 is a circle with equation minus π‘₯ squared minus 𝑦 squared equals minus nine. 𝑃 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 𝑃 and is tangent to the circle 𝐷.

The first thing to notice here is actually we’re 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 𝑦 squared equals π‘Ÿ squared, then the center of the circle if it’s in this form is gonna be equal to zero, zero and the radius of the circle is equal to π‘Ÿ. Okay, great, so we’ve got this now and we’ll actually use this to help us solve the problem.

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

Okay, great, so we’ve now got our equation and actually it’s 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 we’ve got π‘₯ squared plus 𝑦 squared equals three squared. And that’s because if we look back to our general form for the equation of a circle, we’ve got π‘₯ squared plus 𝑦 squared equals π‘Ÿ 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 it’s in the form π‘₯ squared plus 𝑦 squared equals π‘Ÿ 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 𝑦 squared equals π‘Ÿ squared, where π‘Ÿ being the radius. Fab, so we’ve now got that information, let’s move on to the next part of the question.

So what I’ve 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 we’ve got our circle and we’ve got the point 𝑃. And what we’re actually looking for is we actually want to find an equation for the line that passes through the point 𝑃 and is a tangent to the circle at 𝐷. So I’ve drawn that line there. So it’s actually gonna be a tangent to the circle. So what we want to do now is actually find the gradient of our radius 𝑂𝑃. And we want to do that because actually it’s gonna be perpendicular to our tangent. So therefore, it’s gonna allow us to actually work out the gradient of our tangent.

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

Okay, so now, let’s 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 that’s because if we divide by a fraction, that’s the same as multiplying by the reciprocal. So we’re 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 we’re left with minus nine over 12, which we can then simplify to give us our gradient π‘š 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 𝑃. Okay, so now to find the gradient of tangent to 𝐷 touching 𝑃, 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‘š one multiplied by π‘š 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‘š two is equal to minus one. So therefore, π‘š 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 that’s perpendicular to it.

Okay, fab, so now let’s 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 𝑃. Because it’s a straight line, we know it’s gonna be in the form 𝑦 equals π‘šπ‘₯ plus 𝑐. And in this case, we’ve actually put 𝑦 equals π‘š two π‘₯ plus 𝑐 because we already know the gradient π‘š two. So we now know that it’s 𝑦 is equal to four-thirds of π‘₯ plus 𝑐. And as we’ve already said that because we know that four-thirds is equal to π‘š two.

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

So then, what we’re 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 𝑐. So then, we actually add 48 over 15 to both sides. So we get 𝑐 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 𝑐 is equal to 75 over 15, which therefore gives us a final answer of 𝑐 of five.

So therefore, we can say that the equation for the line that passes through the point 𝑃 which is minus 12 over five, nine over five and is tangent to the circle 𝐷 is gonna be 𝑦 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 𝑦-intercept.

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