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Circle question

2022 · 29 Jun · Shift 2 · Q30
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  5. /2022 · 29 Jun · Shift 2 · Q30

Circle question

2022 · 29 Jun · Shift 2 · Q30

JEE MainMathematicsCircleMCQ+4 / −1
Let a triangle ABC be inscribed in the circle x2−2(x+y)+y2=0{x^2} - \sqrt 2 (x + y) + {y^2} = 0x2−2​(x+y)+y2=0 such that ∠BAC=π2\angle BAC = {\pi \over 2}∠BAC=2π​. If the length of side AB is 2\sqrt 22​, then the area of the Δ\DeltaΔ ABC is equal to :
  1. A
    1
  2. B
    (6+3)/2\left( {\sqrt 6 + \sqrt 3 } \right)/2(6​+3​)/2
  3. C
    (3+3)/4\left( {3 + \sqrt 3 } \right)/4(3+3​)/4
  4. D
    (6+23)/4\left( {\sqrt 6 + 2\sqrt 3 } \right)/4(6​+23​)/4
View written solutionFree

Correct answer: A

  1. Identify the circle

Given circle: x2+y2−2(x+y)=0x^2+y^2-\sqrt2(x+y)=0x2+y2−2​(x+y)=0

Compare with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 so 2g=−2,2f=−2,c=02g=-\sqrt2,\quad 2f=-\sqrt2,\quad c=02g=−2​,2f=−2​,c=0

Hence the center is (−g,−f)=(22,22)(-g,-f)=\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)(−g,−f)=(22​​,22​​) and radius is

\sqrt{\left(\frac{\sqrt2}{2}\right)^2+\left(\frac{\sqrt2}{2}\right)^2}=\sqrt{\frac12+\frac12}=1.$$ So the given circle has radius $1$. --- 2. **Use the condition $\angle BAC=\pi/2$** A triangle inscribed in a circle with one angle equal to $90^\circ$ must have the opposite side as the diameter. Therefore, in $\triangle ABC$, since $$\angle BAC=90^\circ,$$ side $BC$ is the diameter of the circumcircle. Hence $$BC=2r=2.$$ --- 3. **Use Pythagoras in the right triangle** Since the triangle is right-angled at $A$, $$AB^2+AC^2=BC^2.$$ Given $$AB=\sqrt2,\quad BC=2,$$ so $$(\sqrt2)^2+AC^2=2^2$$ $$2+AC^2=4$$ $$AC^2=2$$ $$AC=\sqrt2.$$ --- 4. **Find the area** Area of a right triangle: $$\text{Area}=\frac12\cdot AB\cdot AC$$ $$=\frac12\cdot \sqrt2\cdot \sqrt2$$ $$=\frac12\cdot 2=1.$$ --- 5. **Check options** The area is $$1.$$ So the correct option is **A**. --- 6. **Compare with stored answer** Stored correct answer: **A** This matches the derived answer.
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