Gr12 November 2023 finding 'k' in circle transformation | ANALYTICAL GEOMETRY | mathssl

[00:00:00]Welcome to Maths SL. This is November 2023 NSC exam question.

[00:00:08]Concepts about double circles.

[00:00:11]The question says in the diagram the circle with center O has the equation this is the equation.

[00:00:19]And the G is the center of the larger circle.

[00:00:25]So O is the center of small circle and G is the larger circle center.

[00:00:32]A common tangent touches the circle at D and F respectively such that D lies in the fourth quadrant.

[00:00:43]This is the common tangent and the D lies on the fourth quadrant.

[00:00:48]4.1 Given that D P X coordinates P and Y is -2 lies on the smaller circle show that P is equal to 4 to mark question.

[00:01:02]Substitute the coordinates of D into the equation of the smaller circle.

[00:01:10]So by substitution method so this is a X and this is Y so the place of X going to be a P square plus place of Y -2 square is equal to 20.

[00:01:24]Solve for P P square plus 4 is equal to 20.

[00:01:31]Solve for P P square is equal to 20 minus 4 is 16 therefore P is plus minus 4. Since D lies in the fourth quadrant so the X must be greater than zero which means positive so I'm going to take the positive 4 therefore the P is equal to positive 4.

[00:01:52]This is the fourth quadrant.

[00:01:55]Second question E is the midpoint of DF.

[00:01:59]Determine the coordinates of F. Three mark question.

[00:02:03]So, I'm going to use the midpoint formula since E is the midpoint of DF.

[00:02:10]This is the E. So, the D coordinates we have four and negative two we found earlier.

[00:02:17]And the E is given.

[00:02:19]And the E is given six and two. Say F I got X and Y.

[00:02:25]So, I need to find the coordinates of F.

[00:02:28]By midpoint formula, I'm going to solve.

[00:02:32]So, the XE if I need to find So, we're going to say XD plus X F divided by two.

[00:02:45]The XE is given six.

[00:02:48]XD we also got four. XF we need to find.

[00:02:52]Over two, cross multiply.

[00:02:55]So, this is going to be a four plus X.

[00:02:57]Four is equal to 12. So, the X four 12 minus four become eight. So, this is eight.

[00:03:06]So, if I need to find the YE So, YD plus YF divided by two.

[00:03:15]YE given two and YD negative two. YF we need to find divided by two. Cross multiply again.

[00:03:24]YF is equal to four. Therefore, YF is going to be a six.

[00:03:29]There we go.

[00:03:32]So, therefore, coordinates of F are eight and six.

[00:03:38]This is a popular concept they're asking a lot of this.

[00:03:42]4.3 Determine the equation of the common tangent DF in the form Y is equal to MX plus C four mark. So, we're looking DF now.

[00:03:52]So, first of all, I'm going to find the gradient of the radius OD.

[00:03:58]This one.

[00:04:00]So, the gradient of OD is going to be Y2 - Y1 over X2 - X1.

[00:04:05]-2 - 0 over 4 [snorts] - 0. My answer is -1/2.

[00:04:12]Now, I'm going to find the gradient of the tangent perpendicular gradient invert and change the sign. Positive two.

[00:04:20]Now, I'm going to substitute the gradient and the point D into the gradient point form equation.

[00:04:30]This is also popular concept, must concept. So, the Y - - become positive two. The gradient is two.

[00:04:37]X - 4.

[00:04:40]Further simplify.

[00:04:42]2X - 8. Further simplify. 2X - 10.

[00:04:48]So, this is my final answer.

[00:04:52]Calculate the value of T. Show all working. Three mark. Please remember the radius GF perpendicular to the tangent line DF at the point of contact F.

[00:05:08]Coordinates are 8 and 6 we found earlier.

[00:05:11]And the gradient of the tangent we have earlier we found that is two.

[00:05:16]So, the gradient of the radius going to be now, which is GF invert and change the sign. -1/2.

[00:05:24]And I need to find the T here.

[00:05:27]I'm going to find the gradient between GF. I'm going to equate.

[00:05:32]So, it's going to be a 6 - 0 over 8 - T is equal to -1/2. I'm going to equate the gradient.

[00:05:43]6 - 0 over 8 - T is equal to - 1 over 2.

[00:05:47]Further simplify cross multiply. This is going to be a - 8 - - become positive t equal to 12.

[00:05:55]This negative I'm going to pull up in my choice. Can take it to as a numerator or as a denominator.

[00:06:01]I'm using as a numerator.

[00:06:04]So, therefore t is going to be equal to - 8 I pull over this side become positive 20. This is my final answer.

[00:06:14]4.5 determine the equation of the larger circle in the form in this form.

[00:06:19]This is the general form.

[00:06:22]This is the center of the circle we found earlier value of t 20.

[00:06:28]And [snorts] this is the x-axis and you can see this is the x-intercept y value is equal to zero.

[00:06:34]Because this line radius touching the x-axis.

[00:06:40]And point f on the circumference of the circle first need to find the standard circle equation.

[00:06:49]So, the x - 20 squared + y - 0 squared is equal to r squared.

[00:06:59]Now, to find the square radius I'm going to substitute this f point into this equation.

[00:07:06]>> [snorts] >> So, 8 - 20 squared + y + y going to be 6 - 0 squared is equal to r squared.

[00:07:18]So, this is going to be 144 + 36 is equal to r squared.

[00:07:24]180 is equal to r squared. So, the r squared is 180.

[00:07:28]So, the radius squared 180 I'm going to put back into this equation.

[00:07:34]So, x - 20 squared + y - 0 squared is equal to 180.

[00:07:43]Now, I need to convert into the general form expansion. So, I'm going to expand open up the perfect square formula. It's going to be x square - 40 x + 400 + y square this become - 180 is equal to 0.

[00:08:03]Rearrange x square + y square - 40 x + 220 equal to 0.

[00:08:14]So, this is my final answer.

[00:08:16]The last question, what a nice question.

[00:08:19]The smaller circle must be translated by k units along the x-axis to touch the larger circle internally.

[00:08:29]Internally.

[00:08:30]Calculate the possible values of k. Must remember internally.

[00:08:35]Now, must remember for internal touching the distance between the shifted center of the smaller circle and the center of the larger circle must equal the difference between their radii.

[00:08:51]So, this is small r. This is a capital R.

[00:08:55]Must remember the condition d must be equal to big R minus small R then the circle will touch internally.

[00:09:04]So, first I'm looking at this one.

[00:09:07]So, the larger circle R square is equal to 180.

[00:09:10]>> [snorts] >> And if I need to find the radius, it's going to be a square root 180.

[00:09:15]Which mean square root 180 and I use my calculator.

[00:09:21]6 multiply square root 5.

[00:09:26]Smaller circle square radius is equal to 20. If I need to find the R is going to be square root 20 and I use my calculator, my answer is 2 * √5.

[00:09:40]Now, I need to find the difference in radii, big R - small r, and that's going to be a 6 √5 - 2 √5.

[00:09:55]So, my answer is 4 √5.

[00:10:00]Now, the small circle center after translation going to be k and 0 because k unit translated.

[00:10:11]Now, translated along the x-axis, must remember, now, x-axis. [snorts] So, it's going to be x right minus x left.

[00:10:18]So, it's going to be a 20 minus k and it must be equal to 4 √5.

[00:10:25]Remember the condition, d must be equal to big R minus small r, then the circle will touch internally. So, I'm going to use the absolute value bars because distance is always positive.

[00:10:38]20 minus k absolute value is equal to 4 √5.

[00:10:46]So, when you remove the absolute value bars to solve for k, the expressions on the other side must take a positive and negative sign.

[00:10:56]So, it's going to be 20 minus k plus minus 4 multiply 5 √5. Uh now, this accounts for the two possible directions from the center g.

[00:11:09]This is the center g.

[00:11:11]Smaller circle can touch the larger circle internally from two different positions, either on the left side or on the right side of the larger circle center g.

[00:11:24]So, it can touch internally here, and this is the center o here on the right-hand side. See, this is the center.

[00:11:33]And this is the center of the larger circle.

[00:11:37]So, if you want the small circle to touch the inner left wall of the big circle, its center must sit to the left of the of G.

[00:11:48]So, if you want to move left on a graph, you minus the distance from the starting position.

[00:11:55]So, this is a 20 start position minus the distance which is going to be 4 square root 5.

[00:12:04]So, the new center K is going to be equal to 20 minus 4 square root 5. The answer in decimal 11.06.

[00:12:16]I'm solving K here now.

[00:12:19]Now, this mean the circle touch internally left hand side. On the right hand side, if you want small circle to touch the inner right wall of the big circle, its center must sit to the right of the of G.

[00:12:34]Now, you have to move right on the graph, so you must add to the distance.

[00:12:39]So, the Now, again, you're going to solve for K.

[00:12:42]This time I'm going to use positive sign. So, it's going to be K is equal to 20 plus 4 square root 5. My answer is 28.94.

[00:12:53]Same thing here, negative K is going to be equal negative 20 one time negative and one time going to be positive. So, the simplified negative K is equal to negative 28.94.

[00:13:10]Both side you divide by negative, the answer is 28.94. Either way, you can solve this way or that way.

[00:13:16]I just apply a shortcut. That's how you're going to solve K.

[00:13:20]Add the distance touching the right side minus the distance touching the side.

[00:13:27]If the question says the circle translate along any specific line, whether it is the x-axis, y-axis, then there will be two places for internally contact.

[00:13:42]That's why along the x-axis translation and touching left-hand side or can be right-hand side.

[00:13:50]So, simple thing moving along a specific line, going to touch the circle internally two places.

[00:13:57]So, the K is equal to 11.06 or 28.94.

[00:14:01]This is my final answer. Please like and subscribe.