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Maths-

General

Easy

Question

In the $x y$ - plane, the points (-3,10) and (3,10) are endpoints of the diameter of a circle. Which equatic represents this circle?

$x^{2} + (y - 10)^{2} = 9$
$(x + 3)^{2} + y^{2} = 36$
$(x + 3)^{2} + (y - 10)^{2} = 9$
$(x + 3)^{2} + (y - 10)^{2} = 36$

The correct answer is: $x squared plus left parenthesis y minus 10 right parenthesis squared equals 9$

HINT : find the centre of the circle i.e the midpoint of the diameter and find radius of circle and using the centre and radius of circle find the equation of circle .
Explanation
Step 1: find centre of the circle (i.e mid point of diameter)
Given that the endpoints of diameter are (-3,10) and (3,10)
Midpoint of points (x1,y1) and (x2,y2) = [ $fraction numerator left parenthesis x 1 plus x 2 right parenthesis over denominator 2 end fraction comma space fraction numerator left parenthesis y 1 plus y 2 right parenthesis over denominator 2 end fraction$ ]
So the centre is [ $fraction numerator left parenthesis negative 3 plus 3 right parenthesis over denominator 2 end fraction comma space fraction numerator left parenthesis 10 plus 10 right parenthesis over denominator 2 end fraction$ ] [0,10]
Step 2:
Radius of circle = distance between centre (0,10) and point on circle (3,10)
$square root of left parenthesis 3 minus 0 right parenthesis squared plus left parenthesis 10 minus 10 right parenthesis squared end root not stretchy rightwards double arrow r equals 3$
Step 3: The equation of circle with (x1,y1) as centre and r as radius is
$table attributes columnspacing 1em end attributes row cell left parenthesis x minus x 1 right parenthesis squared plus left parenthesis y minus y 1 right parenthesis squared equals r squared end cell row cell left parenthesis x minus 0 right parenthesis squared plus left parenthesis y minus 10 right parenthesis squared equals 3 squared end cell row cell bold italic x squared plus left parenthesis y minus 10 right parenthesis squared equals 9 end cell end table$

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