Mathematics, 31.12.2019 17:31, tanyiawilliams7490

(c) show that the circles x2 + y2 −16x−12y + 75 = 0 and 5x2 + 5y2 −32x−24y + 75 = 0 touch each other.

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$\bf x^2+y^2-16x-12y+75=0
\\\\\\
5x^2+5y^2-32x-24y+75=0$

so, let's do the first one

$\bf x^2+y^2-16x-12y+75=0\impliedby \textit{we'll start by grouping}
\\\\\\
(x^2-16x+\boxed{c}^2)+(y^2-12y+\boxed{d}^2)=-75$

now, in a perfect square trinomial, we know the middle term is just the product of 2 times the square root of the term on the left end, times the square root of the term on the right

that means $\bf (x^2-16x+\boxed{c}^2)+(y^2-12y+\boxed{d}^2)=-75
\\\\\\
2xc=16x\implies c=\cfrac{16x}{2x}\implies \boxed{c}=8
\\\\\\
2yd=12y\implies d=\cfrac{12y}{2y}\implies \boxed{d}=6\\\\
$

so, those are our missing values, now
bear in mind all we're doing is borrowing from our good friend Mr Zero, 0
so, if we add 8² and 6², we also have to subtract 8² and 6²

then $\bf (x^2-16x+8^2)+(y^2-12y+6^2)-8^2-6^2=-75
\\\\\\
(x-8)^2+(y-6)^2=-75+100\implies \boxed{(x-8)^2+(y-6)^2=5^2}$

so, that's the equation of the circle for the first equation, centered at 8,6 and with a radius of 5

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now, onto the 2nd equation, we'll do the same

$\bf 5x^2+5y^2-32x-24y+75=0\impliedby grouping
\\\\\\
(5x^2-32x)+(5y^2-24y)=-75
\\\\\\
5\left( x^2-\frac{32x}{5}+\boxed{e}^2 \right)+5\left( y^2-\frac{24y}{5}+\boxed{f}^2 \right)=-75\\\\
-----------------------------\\\\
2xe=\cfrac{32x}{5}\implies e=\cfrac{16}{5}
\\\\\\
2yf=\cfrac{24y}{5}\implies f=\cfrac{12}{5}\\\\ -----------------------------$

$\bf 5\left( x^2-\frac{32x}{5}+\left( \frac{16}{5} \right)^2 \right)+5\left( y^2-\frac{24y}{5}+\left( \frac{12}{5} \right)^2 \right)-5\left( \frac{16}{5} \right)^2-5\left( \frac{12}{5} \right)^2 
\\\\=-75
\\\\\\
5\left( x-\frac{16}{5} \right)^2+5\left( y-\frac{12}{5} \right)^2=-75+\cfrac{256}{5}+\cfrac{144}{5}$

$\bf 5\left( x-\frac{16}{5} \right)^2+5\left( y-\frac{12}{5} \right)^2=5\implies \left( x-\frac{16}{5} \right)^2+\left( y-\frac{12}{5} \right)^2=\cfrac{5}{5}
\\\\\\
\boxed{\left( x-\frac{16}{5} \right)^2+\left( y-\frac{12}{5} \right)^2=1}$

so, that's a circle centered at 16/5 and 12/5, with a radius of 1
so .those are the two circle's equations

notice, the picture below, the radius of 5, the first equation, is the bigger circle
(c) show that the circles x2 + y2 −16x−12y + 75 = 0 and 5x2 + 5y2 −32x−24y + 75 = 0 touch each other