if the points P(4,3)and Q(x,5)are on the circle with centre O(2,3)then find tje value of x
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OP=OQ (radius)
So,
OP^2=OQ^2
(4-2)^2+(3-3)^2=(x-2)^2+(5-3)^2
4=x^2+4+4x+4
X^2+4x+4=0
X^2+2x+2x+4=0
X(x+2)+2(x+2)=0
X=-2 (Ans)
So,
OP^2=OQ^2
(4-2)^2+(3-3)^2=(x-2)^2+(5-3)^2
4=x^2+4+4x+4
X^2+4x+4=0
X^2+2x+2x+4=0
X(x+2)+2(x+2)=0
X=-2 (Ans)
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