Coordinates of point p which lies on the circle x2 +y2-4x+4y+7=0 in such a way that op is minimum is
Answers
coordinate of point P =(2 - 1/√2 , 1/√2 - 2)
Step-by-step explanation:
x² +y²-4x+4y+7=0
=> (x - 2)² - 4 + (y + 2)² - 4 + 7 =0
=> (x - 2)² + (y + 2)² = 1
=> (x - 2)² + (y + 2)² = 1²
radius = 1
Center ( 2 , - 2)
O is origin (0,0)
Distance between origin & center
= √(2)² + 2² = 2√2
OP = 2√2 - 1
P = ( x , y)
x² + y² = (2√2 - 1 )²
OP would be minimum if OP will lies on line joing center of circle with O
(0.0) , ( 2 , - 2)
Slope = - 1
y = -x + c
=>0 = 0 + c
=> c = 0
=> y = -x
using y = -x
x² + x² = (2√2 - 1 )²
2x² = (2√2 - 1 )²
=> x = (2√2 - 1 )/√2
=> x = 2 - 1/√2
y = -x
=> y = 1/√2 - 2
coordinate of point P =(2 - 1/√2 , 1/√2 - 2)
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