|  | 
| 
 
 
| Question 1085811:  find the equation, in standard form, of the circle being described in each item.
 1.diameter with endpoints (3,-5) & (6,4)
 2.center (-6,-3), through (-10,3)
 Answer by mathmate(429)
      (Show Source): 
You can put this solution on YOUR website! Question: find the equation, in standard form, of the circle being described in each item.
 
 1.diameter with endpoints (3,-5) & (6,4)
 2.center (-6,-3), through (-10,3)
 
 Solution:
 Standard form of a circle is
 (x-x0)^2+(y-y0)^2=r^2
 
 1.
 Diameter end points = (3,-5), (6,4)
 Centre=((3+6)/2, (-5+4)/2))=(4.5,-0.5)
 radius, r = (1/2)sqrt((6-3)^2+(4--5)^2)=(1/2)sqrt(9+81)=(1/2)sqrt(90)=(3/2)sqrt(10),
 r²=(3/2)²(sqrt(10)²)=9/4*10=45/2
 So standard form:
 (x-4.5)²+(y+0.5)²=45/2
 
 2. radius = sqrt((-10--6)²+(3--3)²)=sqrt(16+36)=sqrt(52)
 r²=52
 standard form:
 (x--6)²+(y--3)²=52
 =>
 (x+6)²+(y+3)²=52
 | 
  
 | 
 |  |  |