document.write( "Question 1044110: The centre of a circle has coordinates (0,0) One end of a diameter is located at (7,-2)\r
\n" ); document.write( "\n" ); document.write( "a) What are the coordinates of the other endpoint of this diameter? \r
\n" ); document.write( "\n" ); document.write( "b) What is the equation of the circle?
\n" ); document.write( "

Algebra.Com's Answer #659368 by advanced_Learner(501)\"\" \"About 
You can put this solution on YOUR website!
The centre of a circle has coordinates (0,0) One end of a diameter is located at (7,-2)\r
\n" ); document.write( "\n" ); document.write( "a) What are the coordinates of the other endpoint of this diameter? \r
\n" ); document.write( "\n" ); document.write( "b) What is the equation of the circle? \r
\n" ); document.write( "\n" ); document.write( "many ways to tackle this.
\n" ); document.write( "\"c%280%2C0%29\",\"D%287%2C-2%29\"
\n" ); document.write( "by inspection the other end of the diameter is located at \"D%28-7%2C2%29\".
\n" ); document.write( "use midpoint formula to verify
\n" ); document.write( "(\"0%2C0%29\"=\"%28a%2B7%29%2F%282%29%2B%28b%2B-2%29%2F2\"
\n" ); document.write( "means
\n" ); document.write( "\"%280%29\"=\"%28a%2B7%29%2F%282%29\",\"a%2B7=0\",\"a=-7\"
\n" ); document.write( "\"0\"=\"%28b%2B-2%29%2F2\",\"b-2=0\",\"b=2\" therefore \"D%28-7%2C2%29\"\r
\n" ); document.write( "\n" ); document.write( "two ways
\n" ); document.write( "find the distance from end of the diamater to other end to get diameter ,divide by 2 to get radius.using (7, -2) and (-7, 2)\r
\n" ); document.write( "\n" ); document.write( "\"D+\" = \"%28+sqrt+%28+212%29%29\"
\n" ); document.write( "\"D+\" = \"%28+sqrt+%28+212%29%29\"
\n" ); document.write( "\"D+\" = 2\"%28+sqrt+%28+53%29%29\"
\n" ); document.write( "half of it is the radius or \"R+\" = \"%28+sqrt+%28+53%29%29\"\r
\n" ); document.write( "\n" ); document.write( "same can be found using origin and one end of the diameter to get radius.
\n" ); document.write( "distance from one end to the origin c(0,0) to get radius using ((0 ,0))and ((7 ,-2))
\n" ); document.write( "\"R\" = \"%28+sqrt+%28+53%29%29\"\r
\n" ); document.write( "\n" ); document.write( "the equation is
\n" ); document.write( "\"x%5E2%2By%5E2\"=\"r%5E2\"
\n" ); document.write( "\"x%5E2%2By%5E2\"=\"%28sqrt%2853%29%29%5E2%0D%0A%0D%0A%7B%7B%7Bx%5E2%2By%5E2\"=\"53\"
\n" ); document.write( "
\n" );