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A circle having its centre in the first quadrant touches the y axis at point (0,2) and passes through the point (1,0) . find the equation of circle.
CONSIDER AN ARBITARY EQUATION OF CIRCLE WITH CENTER (-g,-f) OR IN OTHER WORDS IF YOU HAVE STUDIED THREE POINT FORM IN CIRCLE THEN EQUATION IS X2+Y2+2gX+2fY+C=0 SINCE WE KNOW THAT IT PASSES THROUGH (0,2) AND (1,0) SUBSTITUTE THESE VALUES IN THE EQUATION OF THE CIRCLE INSTEAD OF X AND Y SINCE IT SATISFIES THE EQUATION OF THE CIRCLE YOU WILL GET TWO EQUATION BUT THREE VARIABLES CONTAINING g,f,c NOW CLEARLY FROM THE DIAGRAM WE CAN SEE THAT THE VALUE OF -f=2 therefore f = -2 THEN SUBSTITUTE THE VALU OF f IN THE EQUATION CONTAINING g,f,c AS VARIABLES. NOW YOU WILL GET TWO EQUATION IN TWO VARIABLES WHICH IS g,c SOLVE FOR g,c AND THEN SUBSTITUTE THE VALUE OF g,f,c IN THE EQUATIN OF CIRCLE AND GET YOUR ANSWER. IF YOU ARE SATISFIED WITH THE ANSWER PLZZ DO APPROVE IT
CONSIDER AN ARBITARY EQUATION OF CIRCLE WITH CENTER (-g,-f) OR IN OTHER WORDS IF YOU HAVE STUDIED THREE POINT FORM IN CIRCLE THEN EQUATION IS
X2+Y2+2gX+2fY+C=0 SINCE WE KNOW THAT IT PASSES THROUGH (0,2) AND (1,0) SUBSTITUTE THESE VALUES IN THE EQUATION OF THE CIRCLE INSTEAD OF X AND Y SINCE IT SATISFIES THE EQUATION OF THE CIRCLE YOU WILL GET TWO EQUATION BUT THREE VARIABLES CONTAINING g,f,c NOW CLEARLY FROM THE DIAGRAM WE CAN SEE THAT THE VALUE OF
-f=2 therefore f = -2 THEN SUBSTITUTE THE VALU OF f IN THE EQUATION CONTAINING g,f,c AS VARIABLES. NOW YOU WILL GET TWO EQUATION IN TWO VARIABLES WHICH IS g,c SOLVE FOR g,c AND THEN SUBSTITUTE THE VALUE OF g,f,c IN THE EQUATIN OF CIRCLE AND GET YOUR ANSWER.
IF YOU ARE SATISFIED WITH THE ANSWER PLZZ DO APPROVE IT
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