Video Transcript
Find the centre and radius of the circle ๐ฅ squared plus ๐ฆ squared plus 10๐ฅ minus 18๐ฆ plus 42 equals zero.
So weโve been given the equation of a circle in its expanded form, and weโre asked to find its centre and radius. In order to do so, we need to rearrange this equation so that we have it in the standard form for the equation of a circle, the centre-radius form. This standard form is ๐ฅ minus โ squared plus ๐ฆ minus ๐ squared equals ๐ squared, where the centre of the circle is the point with coordinates โ, ๐ and the radius of the circle is ๐.
In order to rearrange the equation of the circle into this form, we need to complete the square. First, Iโm going to reorder the terms so that the ๐ฅ squared and the ๐ฅ terms are together, the ๐ฆ squared and the ๐ฆ terms are together, and then Iโve subtracted 42 from both sides. So I now have negative 42 on the right-hand side of the equation.
So I am going to complete the square in ๐ฅ first of all. I need to halve the coefficient of ๐ฅ, so thatโs 10 and square this value. So half of 10 is five, and when I square this, I get 25. However, adding this 25 in, I have unbalanced my equation. So I need to add it onto the right-hand side as well.
Now we complete the square in ๐ฆ. So we look at the coefficient of ๐ฆ which is negative 18. Halving that, we have negative nine and then squaring that, we have 81. Again, if I only add this to one side of the equation, then Iโve unbalanced it. So I also need to add 81 to the right-hand side.
Now looking at the first three terms, we can see that we have a perfect square. ๐ฅ squared plus 10๐ฅ plus 25 is equivalent to ๐ฅ plus five all squared. And thatโs not a coincidence, thatโs by design because weโve been using the completing the square method. So we replace the first three terms with ๐ฅ plus five all squared.
Now letโs look at the next lot of terms, those in ๐ฆ. We have ๐ฆ squared minus 18๐ฆ plus 81, which is also a perfect square. Itโs equal to ๐ฆ minus nine all squared. So we replace these next two terms with ๐ฆ minus nine all squared. On the right-hand side, we have negative 42 plus 25 plus 81 which simplifies to 64.
Now if we compare the equation of our circle with the standard form, we can see that weโre nearly there. The value of ๐, in particular, we can see is nine. If we look at the ๐ฅ part, then we have ๐ฅ plus five all squared, and weโre comparing this with ๐ฅ minus โ all squared. This means that we must be subtracting negative five in order to reach ๐ฅ plus five. So we have ๐ฅ minus negative five all squared. And now we can see the value of โ as well.
Finally, letโs look at the value of ๐. On the right-hand side of the equation, we have 64. And 64 is equal to โ squared. So now if we compare this with the standard form, we can see that ๐ is equal to eight.
So letโs put all this together then and write down the centre and the radius of the circle. Weโve found that the centre of the circle is the point with coordinates negative five, nine and the radius of the circle is eight length units.