Quantitative Aptitude – Geometry – The vertices of a triangle are (0,0)
Slot – 3 – Quantitative Aptitude – Geometry – The vertices of a triangle are (0,0)
Q. The vertices of a triangle are (0,0), (4,0), (3,9). The area of the circle passing through these three points is?
127π/7
205π/9
14π/3
12π/5
Answer: 205π/9
Solutions:
The equation of circle is
(x-h)^2+ (y-k)^2=r^2
All these points will satisfy this equation
For (0,0) , Put x= 0, y= 0
h^2+k^2=r^2 —…………..1)
For (4,0) , Put x= 4, y= 0
(4-h)^2+k^2=r^2 ………………….2)
For (3,9) , Put x= 3, y= 9
(3-h)^2+(9-k)^2=r^2 …………..3)
From eq 1) and 2)
(4-h)^2+r^2-h^2=r^2
Or h = 0 or 2
By eq 3 – eq 1
9 – 6h + 81 – 18k =0
90 – 18k = 6h
If h = 0
k = 5 and if h =2 , k = 13/3
thus from eq 1)
if (h,k) = (0,5) -> r = 5
Area of circle = πr^2=25π
If (h,k) = (2,17/3) -> r^2=4+169/9=205/9
Area of circle = πr^2=205π/9
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