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Question
If the vertices of a triangle be (2, –2), (–1, –1) and (5, 2), then the equation of its circum circle is
If the vertices of a triangle be (2, –2), (–1, –1) and (5, 2), then the equation of its circum circle is
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Solution
The correct option is B
Let us find the equation of family of circles through (2, -2) and (-1, -1).
i.e. (x−2)(x+1)+(y+2)(y+1)+λ(y+2–2+1−x−22+1)=0
Now for point(5, 2) to lie on it, we should have λ given by
3.6+4.3+λ(4−1−1)=0⇒λ=305=6
Hence equation is
(x−2)(x+1)+(y+2)(y+1)+6(y+2–1−x−23)=0
Or x2+y2−3x−3y−8=0.
Trick: Here the circle x2+y2−3x−3y−8=0 is satisfied by (2, -2), (-1, -1) and (5, 2).
Therefore students must check such type of problems conversely.
Let us find the equation of family of circles through (2, -2) and (-1, -1).
i.e. (x−2)(x+1)+(y+2)(y+1)+λ(y+2–2+1−x−22+1)=0
Now for point(5, 2) to lie on it, we should have λ given by
3.6+4.3+λ(4−1−1)=0⇒λ=305=6
Hence equation is
(x−2)(x+1)+(y+2)(y+1)+6(y+2–1−x−23)=0
Or x2+y2−3x−3y−8=0.
Trick: Here the circle x2+y2−3x−3y−8=0 is satisfied by (2, -2), (-1, -1) and (5, 2).
Therefore students must check such type of problems conversely.
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