Solve each problem. Find all values of b such that the straight l...

Question

Solve each problem. Find all values of b such that the straight line 3x - y = b touches the circle x^2 + y^2 = 25 at only one point.

Accepted Answer

Hey, everyone in this problem, we have the straight line given by negative two X plus Y is equal to C that touches the circle given by X squared plus Y squared is equal to 36. At only one point we're asked to determine all the possible values of C. We're given four answer choices. Option A is six multiplied by the square root of five. Option B negative six multiplied by the square root of five. Option C is both A and B and option D is plus or minus the square root of five. Let's start by writing out the equations. We were given, we have negative two X plus Y is equal to C. We have X squared plus Y squared is equal to 36. Now we're told that these intersect the only one point, OK. In order to find the intersection, what we really have here is two equations. A system of two equations. We're gonna use the substitution method. OK. So we're gonna start with the equation of the line and we're gonna isolate for Y. OK. Now you could isolate for either variable. But if we isolate for X, we're gonna have to divide by negative two and we get some fractions and that's just gonna be a little bit messier to deal with. So we're gonna isolate for Y and doing that, we're gonna move the negative two X to the right by adding two X, we have that Y is equal to C plus two X. We're gonna call this equation one and we're gonna move over to our equation for the circle X squared plus Y squared equals 36. And we're gonna substitute in this equation. One, Ky is equal to C plus two X. What that gives us is X squared plus Y squared, which is now C plus two X squared is equal to 36. OK. So we're just using our substitute method just like we do when we have two linear equations, it's the exact same process. We're just dealing with one of these equations that sets a circle. And now you'll see that our equation has only XS. OK. We've eliminated the Y from our equation. So we have XS and then we have this constant of C that we're trying to find the value of. Let's go ahead and expand and simplify in any way we can, we have X squared plus C plus two X all squared and expanding. That is gonna give us C squared plus four CX plus four X squared and this is all equal to 36. So what you'll notice is we have X squared terms we have X terms and we have these constant terms. OK. Remember that C is just a constant that we don't know the value of yet. So C squared is a constant and 36 is a constant. So we have a quadratic equation for X. So let's move everything to the left hand side and have this equal to zero. We have X squared plus four X squared gives us five X squared. OK. Then we have plus four CX plus plus C squared minus and this is all equal to zero. Now, in order to solve for X and we're gonna use a quadratic formula. And let's recall where the values come from. OK. The A value form our quadratic formula is gonna be the coefficient of our X squared term which is five, the B term is the coefficient of our X term which is four C and the coefficient of our constant term or just our constant term is going to be our C value. Now, I don't want you to mix up the C from our quadratic formula with the C from this equation, right? So when I write this equation in general X is equal to negative V plus or minus the square root of B squared minus four AC divided by two A, the C here is that general C which is that entire constant term. OK. Once we substitute in the values, we're gonna be talking specifically about the C that was given in our linear equation. Now, we have our quadratic formula and we can substitute in our values and solve for this what the X value we get. But what we wanna do is use the information we were given in the problem. And we were told in the problem that this has only one solution. Now, in order to have only one solution, what we need is everything the square root to be zero. If this is zero, then we get negative B plus or minus zero. Whether we add zero or subtract zero, we get the same answer. And so we only end up with one X value at negative B divided by two A. OK. So for only one solution, we want B squared minus four AC everything under that square root to be equal to zero, substituting in our values. OK. B is gonna be four C. So we have four C squared minus four multiplied by A which is five multiplied by our C value which is C squared minus 36. And all of this is equal to zero. And now that we've substituted in our values, the C in this equation is that C from our linear equation, we are no longer talking about the general C or our quadratic formula. Let's go ahead and simplify. You can see now that we have an equation where the only unknown is C. This is gonna allow us to solve for that value. Of C. So we have four C all squared that gives us 16 C squared, we have negative 20 multiplied by C squared minus 36. This gives us negative 20 C squared, 720. This is equal to zero C squared minus 20 C squared is going to give us negative four C squared. We're gonna move that to the right hand side. OK by adding and so we end up with four, C squared is equal to dividing both sides by four to isolate. For C, we have C squared is equal to 108. And so C is gonna be equal to positive or negative the square root of 108. And we have to take that plus or minus when we take the square root. Now we have the square root of 180. This is not a perfect square. So what we wanna do is try to write 180 in terms of factors that are perfect squares. And it turns out we can write 180 as the square root of 36 multiplied by five A, the square root of 180 is the square root of 36 multiplied by five, the, of 36. We know that that's equal to six. So we get C is equal to plus or minus six, multiplied by the square root of five. And these are possible C values. So even though we were told that we have only one intersection between this line in the circle, we still get two possible C values that will give us that result. OK. So if we go back up to our answer traces and we look at what we were given. OK. We found that the correct answer is option C positive or negative six multiplied by the square root of five. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each problem. Find all values of b such that the straight line 3x - y = b touches the circle x^2 + y^2 = 25 at only one point.

Key Concepts

Circle Equation

Line Equation

Tangency Condition

Watch next