Test Index

Circles Part 2

Section: Mathematics

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Question : 76 of 87

Marks: +1, -0

Let a triangle

ABC

be inscribed in the circle

x^2-\sqrt{2}(x+y)+y^2=0

\angle BAC = \frac{\pi}{2}

. If the length of side

A B

\sqrt{2}

, then the area of the

\triangle ABC

[29-Jun-2022-Shift-2]

1
$\frac{\sqrt{6}+\sqrt{3}}{2}$
$\frac{3+\sqrt{3}}{4}$
$\frac{\sqrt{6}+2\sqrt{3}}{4}$

Solution: 👈: Video Solution

For equation of circle

x^2+y^2+2gx+2fy+c=0

, center is

(-g,-f)

and radius

r=\sqrt{g^2+f^2-c}

Given,

equation of circle is

x^2-\sqrt{2}(x+y)+y^2=0

\Rightarrow x^2+y^2-\sqrt{2}x-\sqrt{2}y=0

\Rightarrow x^2+y^2+2\left(-\frac{1}{\sqrt{2}}\right)x+2\left(-\frac{1}{\sqrt{2}}\right)=0

\therefore g=-\frac{1}{\sqrt{2}} \text{ and } f=-\frac{1}{\sqrt{2}}

\therefore \text{Center}=(-g,-f)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)

\text{And Radius}=r=\sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2-0}

=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}=1

As

A B

and

A C

makes an angle

90^{\circ}

then line

B C

passes through the center of circle and

B C

is the diameter of the circle.

\therefore

Length of

B C=2r=2\times 1=2

\therefore AC^2 = BC^2 - AB^2

=2^2-(\sqrt{2})^2

=2

\Rightarrow AC = \sqrt{2}

\therefore

Area of right angle triangle

ABC

= \frac{1}{2} \times AC \times AB

= \frac{1}{2} \times \sqrt{2} \sqrt{2}

=1 \text{ square unit.}

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