a circle with centre as origin passes through a point(5/2,0). does point (5/2,5/2) loes in its interior or exterior?
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A circle with centre as origin passes through a point (5/2, 0)
so, equation of circle is (x - 0)² + (y -0)² = (5/2-0)² + (0-0)²
⇒x² + y² = 25/4 is the equation of circle passes through (5/2,0) and centre of it is origin.
now put the point ( 5/2,5/2) in equation of circle ,S(x, y) = x² + y² - 25/4 = 0
S(x, y) = (5/2)² + (5/2)² - 25/4
= 25/4 + 25/4 - 25/4 = 25/4
Hence, S(x,y) = 25/4 > 0 , it means point (5/2,5/2) lies outside the circle.
[ note : if any circle S(x,y) is given and point (x₁,y₁) are also given .
If S(x₁,y₁) > 0 , then point lies outside the circle
If S(x₁,y₁) < 0 , then point lies inside the circle
if S(x₁,y₁) = 0 ,then point lies on the circle ]
so, equation of circle is (x - 0)² + (y -0)² = (5/2-0)² + (0-0)²
⇒x² + y² = 25/4 is the equation of circle passes through (5/2,0) and centre of it is origin.
now put the point ( 5/2,5/2) in equation of circle ,S(x, y) = x² + y² - 25/4 = 0
S(x, y) = (5/2)² + (5/2)² - 25/4
= 25/4 + 25/4 - 25/4 = 25/4
Hence, S(x,y) = 25/4 > 0 , it means point (5/2,5/2) lies outside the circle.
[ note : if any circle S(x,y) is given and point (x₁,y₁) are also given .
If S(x₁,y₁) > 0 , then point lies outside the circle
If S(x₁,y₁) < 0 , then point lies inside the circle
if S(x₁,y₁) = 0 ,then point lies on the circle ]
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