Question

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

Answer 1

`\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-(32x)/(5)+\boxed{e}^2 \right)+5\left( y^2-(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-(32x)/(5)+\left( (16)/(5) \right)^2 \right)+5\left( y^2-(24y)/(5)+\left( (12)/(5) \right)^2 \right)-5\left( (16)/(5) \right)^2-5\left( (12)/(5) \right)^2 \\\\=-75 \\\\\\ 5\left( x-(16)/(5) \right)^2+5\left( y-(12)/(5) \right)^2=-75+\cfrac{256}{5}+\cfrac{144}{5}`


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