In Exercises 67–70, graph both equations in the same rectangular ...

Question

In Exercises 67–70, graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. (x − 2)²+(y+3)² = 4, y = x - 3

Accepted Answer

Welcome back. I am so glad you're here. We are given two equations. The first one is parentheses X plus one and parentheses squared plus new parentheses y minus three and parentheses squared equals nine. The second equation we're given is Y equals X plus one. And we need to do two things. We need to graph both equations and we need to determine their points of intersection. Well, in order to do that, I am going to first give us some more room and once we've got more room we'll start by graphing the equation on the left. The X plus one squared plus y minus three squared equals nine. We look at this equation and we recall from previous lessons that this equation is the equation of a circle in its standard form which is parentheses X minus H squared plus new parentheses Y minus K squared equals r squared where h comma K. Is the center of that circle and r is the radius of the circle. And with that information the center and the radius that's enough for us to graph it. So let's identify R h R H comes from this X plus one but it's supposed to be x minus h. Well where's r minus h X plus one is the same as x minus a negative one. And there's our x minus h. R h is negative one. So the center for this circle, the H value is negative one. How about R k? Well we see that our k is being subtracted from the Y. Well this one's y minus three. So R k is three. There's our center now how about the radius well? We know that R squared equals two equals nine. And our is our radius. So if we set our squared equal to nine, if we want to solve for R we can take the square root of both sides. And on the left we've got our our which is what we want on the right we have plus or minus the square root of nine which is three plus or minus three. Well we can't have a negative radius so we can reject the negative choice here and it's just a radius of a positive three. And that's enough information to graph this. I typically I will draw the roughest of sketches of a coordinate plane. But this time I'm actually going to try to draw it as precisely as I can. So it'll take a little bit longer. We've got a horizontal X axis and the vertical Y axis. They come together the origin in the middle. From the origin we're going to go to the right marking our X. Units. So we've got 12345. Moving to the right on the X axis and from the origin now moving left we have negative one negative two negative three negative four and negative five. Now from the Origin going up for our Y values we have positive 12345. Mark that as five on the Y axis. And from the origin going down for our Y values negative one negative two negative three negative four and negative five. Now we want to graph this as precisely as we can. And so for our center of negative 13 starting at the origin we're moving to the left one unit for our X. Value of negative one. And we're going up three units for our Y. Value of three. And that's approximately here for the radius of three. That means that anywhere three units away from our center that's our circle. I'm going to mark the ones that are going up down left and right. So from our center going up three units that's up 123. That's right here. So I'm going to mark this. Let me go back and mark the center real quick. So that's negative 13. And marking this one that's three units up. That is going to be still negative one. And now six for the Y value. Now going to the right three units. That's 123 right here from our center And that's going to be at 23. Now from our center going down three units 123, that's right here on the X axis and that's at the point negative 10. And from the center going to the left three units that's 123 right here and that's going to be at -43. We can connect these dots. And I mean this is going to be a pretty poor circle but that is approximately where our circle is. And next we've got that one graft. We can now graph our second equation. Y equals X plus one. Well, we take a look at that equation and we recognize that is the equation of a line and specifically a line in slope intercept form Y equals M. X plus B. Where M. Is our slope and B. Is our Y intercept. And again, that's enough information to graph this our M our slope that's in front of the X. This time our slope is one. Or we could write that as 1/1 and it's actually also the same as negative one over negative one. So we'll use those in a second. R B r y intercept. That's here on the end. That is a positive one. So that's saying when X equals zero Y equals one. Now let's graph that point were from the origin. We're going to graph the 10.0.1. So we're going nowhere for our X values and we're going up one unit for our Y value of one. That's right here, mark this at 01. Now for our slope rise over run we're going up one unit into the right one unit for our Y and X values. That's right here, we can continue going up one unit to the right one unit. That is right here at 23. They both have an intersection there. We could continue going up to the up one unit into the right one unit. We can also from our Y intercept, go in the other direction. We can go down one unit and to the left one unit that's that negative one over negative one down one unit into the left one unit that puts us right here at negative 10. So our line has an intersection there as well. We could connect these dots and yes it would go right through that where I wrote -10. But I'm just going to leave that because we don't want to lose track of that information. So please know that it does go through there and this is our Y. Equals X plus one. We have graphed this and we found our points of intersection. The points of intersection are at negative + and at +23. That's all this was asking for. I will go up and show the appropriate answer choice. But then if you're sitting here thinking, okay, well this is great. I'm glad I can see it. But what if this was shifted a little bit and it wasn't clear at all where the points of intersection are. What if it wasn't right where I knew it was? What if it was just somewhere off to the side like this? And I couldn't just figure that out. Well, I will show you a way to solve for that after if you're interested but to keep this as short as possible for someone who just needed this information, let's go look for our answer. So we'll drag this up and we're looking for points of intersection at -1, 0 and 2 3. I'm kind of blocking that a little bit but we can see from looking at these answer choices that the appropriate one, points of intersection at negative 10 and 23 is answer choice. A Well done. Now if you're interested in seeing how to do this when it's not straightforward, stick with me. I'll show you how to solve this algebraic lee and it won't take too long. Alright bringing this back down now, how do we solve this when it's not straightforward? So what you do is you take your original equations. So we've got the equation of the circle, X plus one squared plus y minus three squared equals nine. And your equation for the line, Y equals x plus one. That's a strange plus X plus one. And now you can use substitution, you know that Y equals X plus one. Well, you can throw x plus one in for y in the equation of the circle. So that would be X plus one squared plus instead of y minus three, it will be x plus one minus three squared equals nine. And now we can write this out, X plus one squared is x plus one times X plus one. And with this X plus one minus three. Well one minus three is negative two. So this will be an x minus two times X minus two equals nine. And now we can foil all of these. So our first x times X is X squared outside X times one is plus one X. Or just plus X insides one times X plus X. Last one times one is plus one plus doing the same thing here foiling this out. First X times X is X squared outside X times negative two is minus two. X inside's negative two times X minus two X. And last negative two times negative two is plus four still equal to nine. Now combining like terms X squared plus X squared gives us two X squared our exes. We have one X two X minus two X. S. Those are gone now just minus two more exes. And then we have plus one plus four, That's plus five equals nine. Let's subtract nine from both sides. That this is equal to zero. Then we've got two X squared minus two. X plus five minus nine is minus four equals zero. Now we can divide everything here by two because it's all divisible by two. And then we're going to get X squared minus X minus two equals zero. We can factor this. And what times what gives us X squared. That's X times X. What? Two numbers when multiply give us a negative two but add up to a negative one. X. That's going to be a minus two plus one. We set both of these factors equal to zero and we can add two to both sides. For our x minus two equals zero. That gives us an X value of two. And for X plus one equals zero. We can subtract one from both sides and that gives us X equals negative one. Look at this, X equals two. That's our X value for the intersection and X equals negative one. That's the other X value for our intersection. Now we can plug those back into the original equation and we'll get the corresponding y values for each of those points and note that we can use the equation of Y equals X plus one because that is shorter. So instead of Y equals X plus one, it'll be Y equals two plus one and two plus one is three. So where X equals to Y equals three. That's exactly what we had. And will do the same thing. Why equals instead of X plus one? Is going to be a negative one plus one, negative one plus one equals zero. So where X equals negative one, Y equals zero. And that's how you solve it algebraic lee if this thing were shifted and not so straightforward and easy to see from the graph. Well done. We'll catch you on the next one

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In Exercises 67–70, graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. (x − 2)²+(y+3)² = 4, y = x - 3