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x-3-2-y-5-2-45-A-0-11-of-the-circle-to-the-point-of-trying-to-find-the-angular-coefficient-




Question Number 12968 by @ANTARES_VY last updated on 08/May/17
(x+3)^2 +(y−5)^2 =45   A(0;11)  of  the  circle  to  the  point  of  trying  to  find the  angular  coefficient.
(x+3)2+(y5)2=45A(0;11)ofthecircletothepointoftryingtofindtheangularcoefficient.
Answered by 433 last updated on 08/May/17
(X,Y)=(x+3,y−5)  X^2 +Y^2 =45 &A(3,6)  X_0 X+Y_0 Y=r^2   3X+6Y=45  3(x+3)+6(y−5)=45  3x+6y=66  y=−(1/2)x+11  I hope i help
(X,Y)=(x+3,y5)X2+Y2=45&A(3,6)X0X+Y0Y=r23X+6Y=453(x+3)+6(y5)=453x+6y=66y=12x+11Ihopeihelp

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