Respuesta :
Answer:
a) [tex](-25,0)\, ,r=1[/tex]
b) [tex](-1,4)\, ,r=3[/tex]
c) [tex](10,5)\, ,r=10[/tex]
d) [tex](0,0)\, ,r=\sqrt{19}[/tex]
e) [tex](-\frac{1}{2},-\frac{1}{2})\, ,r = \dfrac{1}{2}[/tex]
Step-by-step explanation:
the general equation of a circle is can be represented in two forms
- [tex](x-a)^2+(y-b)^2=r^2[/tex]
here,
the centre is: [tex](a,b)[/tex]
radius is: [tex]r[/tex]
- [tex]x^{2}+y^{2}+2gx+2fy+c=0[/tex]
here,
the centre is: [tex](-g,-f)[/tex]
radius is: [tex]\sqrt{g^2 + f^2 -c}[/tex]
Given these two forms we can solve the questions:
a) [tex](x+25)^2+y^2=1[/tex]
this resembles the first equation of the circle hence we can compare them.
for center: we can see that +25 is the place of '-a', there is nothing in the place of b.
[tex]-a = 25[/tex] solve for a
[tex]a = -25[/tex]
similarly,
[tex]b = 0[/tex]
the center is [tex](a,b) = (-25,0)[/tex]
for radius:
[tex]r^2 = 1[/tex]
[tex]r = \sqrt{1}[/tex]
[tex]r = 1[/tex], we'll not take '-1' since that negative numbers don't apply to lengths.
b) [tex]x^2+2x+y^2-8y=8[/tex]
we can clearly see that this resembles the second general equation. Hence by comparison we can find +2x is in the place of +2gx, and solve for g.
[tex]+2gx=+2x[/tex]
[tex]g=1[/tex]
similarly
[tex]+2fy=-8y[/tex]
[tex]f=-4[/tex]
the centre of the circle is denoted by [tex](-g,-f) = (-1,4)[/tex]
for radius:
[tex]r = \sqrt{g^2 + f^2 -c}[/tex]
side note: the value of 'c' can be found if we simply move 8 from the right hand side of the equation to the left. [tex]x^2+2x+y^2-8y-8=0[/tex]
[tex]r = \sqrt{1^2 + (-4)^2 -(-8)}[/tex]
[tex]r = \sqrt{9}[/tex]
[tex]r = 3[/tex]
c) [tex]x^2-20x+y^2-10y+25=0[/tex]
we'll do these in the same manner as (b)
[tex]+2gx=-20x[/tex]
[tex]g=-10[/tex]
[tex]+2fy=-10y[/tex]
[tex]f=-5[/tex]
the centre of the circle is denoted by [tex](-g,-f) = (10,5)[/tex]
for radius:
[tex]r = \sqrt{g^2 + f^2 -c}[/tex]
[tex]r = \sqrt{(-10)^2 + (-5)^2 -25}[/tex]
[tex]r = \sqrt{(-10)^2 + (-5)^2 -25}[/tex]
[tex]r = 10[/tex]
d) [tex]x^2+y^2=19[/tex]
this is easy in the sense that both general equations apply here. We can either say since there is no 2gx and 2fy term, both g and f are zeros hence the centre is (0,0)
or we can also say there's no 'a' and 'b', hence the centre is (0,0).
the centre is: [tex](a,b) = (-g,-f) = (0,0)[/tex]
the radius can also be found through any of the two equations, but to keep it short we'll just use:
[tex]r^2 = 19[/tex]
[tex]r = \sqrt(19)[/tex]
e) [tex]x^2+x+y^2+y=-1/4[/tex]
this is one is again similar to the second equation.
for centre:
[tex]+2gx = x[/tex]
[tex]g = \dfrac{1}{2}[/tex]
[tex]+2fy = y[/tex]
[tex]f = \dfrac{1}{2}[/tex]
the centre of the circle is denoted by [tex](-g,-f) = (-\frac{1}{2},-\frac{1}{2})[/tex]
for radius:
[tex]r = \sqrt{g^2 + f^2 -c}[/tex]
[tex]r = \sqrt{\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 -\left(\dfrac{1}{4}\right)}[/tex]
[tex]r = \sqrt{\dfrac{1}{4}\right}[/tex]
[tex]r = \dfrac{1}{2}[/tex]
