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Sagot :
Answer:
A common form to write the equation of a circle in is the center-radius form. The center-radius form is: (x-h)2+(y-k)2=r2+ , the center point is denoted by (h,k) and r is the
radius of the circle.
Answer:
In order to find the center and radius, we need to change the equation of the circle into standard form,
( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 (x−h)2+(y−k)2=r2,
where h and k are the coordinates of the center and r is the radius.
What is the standard form for the equation of a circle?
In this lesson we’ll look at how to write the equation of a circle in standard form in order to find the center and radius of the circle.
The standard form for the equation of a circle is
(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2
where rr is the radius and (h,k)(h,k) is the center.
Sometimes in order to write the equation of a circle in standard form, you’ll need to complete the square twice, once for xx and once for yy.
Example
Find the center and radius of the circle.
x^2+y^2+24x+10y+160=0
In order to find the center and radius, we need to change the equation of the circle into standard form, (x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2, where hh and kk are the coordinates of the center and rr is the radius.
In order to get the equation into standard form, we have to complete the square with respect to both variables.
Grouping xx’s and yy’s together and moving the constant to the right side, we get
(x^2+24x)+(y^2+10y)=-160
Completing the square requires us to take the coefficient on the first degree terms, divide them by 22, and then square the result before adding the result back to both sides.
The coefficient on the xx term is 2424, so
\frac{24}{2}=12224=12
12^2=144122=144
The coefficient on the yy term is
In order to find the center and radius, we need to change the equation of the circle into standard form, ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 (x−h)2+(y−k)2=r2, where h and k are the coordinates of the center and r is the radius.
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