Answer:
1. \((x+5)^2+(y+7)^2=36\)
2. \((x-5)^2+y^2=17\)
3.
\(r=4\\\\(h,k)=(5,-3)\)
4.
\((2,-1)\\\\and\\\\(2,-5)\)
Step-by-step explanation:
A circle is the set of all points in a plane at a given distance (radius) from a given point (center). The standard equation of a circle with center (h,k) and radius r is given by:
\((x-h)^2+(y-k)^2=r^2\)
Therefore, we just need to use the previous equation in order to solve the problems.
1. Replacing the data provided into the equation of the circle:
\((x-(-5))^2+(y-(-7))^2=6^2\\\\(x+5)^2+(y+7)^2=36\)
2.
\((h,k)=(5,0)\\(x_o,y_o)=(1,1)\)
In this case, we have the center of the circle and a point of the circle, so, let's use that information in order to find the radius:
\((1-5)^2+(1-0)^2=r^2\\\\(-4)^2+(1)^2=r^2\\\\17=r^2\\\\r=\sqrt{17}\)
Hence, the equation of a circle with center (5, 0) that passes through the point (1, 1) is:
\((x-5)^2+(y-0)^2=(\sqrt{17} )^2\\\\(x-5)^2+y^2=17\)
3. We can extract the solution directly from the equation:
\((x-5)^2+(y+3)^2=16\)
The radius is:
\(r^2=16\\\\r=\sqrt{16} \\\\r=4\)
And the center is:
\(x-h=x-5\\\\Solving\hspace{3}for\hspace{3}h\\\\-h=-5\\\\h=5\\\\y-k=y+3\\\\Solving\hspace{3}for\hspace{3}k\\\\-k=3\\\\k=-3\)
So:
\(Center=(h,k)=(5,-3)\)
4. We can find one of the points on this circle:
\((x-2)^2+(y+3)^2=4\)
Simply, eliminating one of the variables. For example, let's elimate x evaluating the equation for x=2:
\((2-2)^2+(y+3)^2=4\\\\(0)^2+y^2+6y+9=4\\\\y^2+6y+5=0\)
Solving for y:
\((y+5)(y+1)=0\)
Hence:
\(y=-1\\\\or\\\\y=-5\)
So, two points that are on the circle are:
\((2,-1)\\\\and\\\\(2,-5)\)
Let's verify it evaluating the points into the equation of this circle.
For (2, -1)
\((2-2)^2+(-1+3)^2=4\\\\0+(2)^2=4\\\\4=4\)
And for (2, -5)
\((2-2)^2+(-5+3)^2=4\\\\0+(-2)^2=4\\\\4=4\)
Since they satisfy the equation we can conclude that those points are on the circle.
Additionally I leave you the graph of every circle.
