Wait, why is $BC$ the diameter? Can't it be any chord?

The theorem states that an angle of $90^\circ$ inscribed in a circle must subtend a diameter. If the angle were not $90^\circ$, $BC$ would not be the diameter. Since $\angle BAC = 90^\circ$, $BC$ must pass through the center.

Why do we complete the square?

The general form $x^2 + y^2 - \sqrt{2}x - \sqrt{2}y = 0$ hides the center and radius. Completing the square transforms it into $(x - \frac{1}{\sqrt{2}})^2 + (y - \frac{1}{\sqrt{2}})^2 = 1$, making the geometry transparent.

Is there a faster way to find the area?

Once you know it's an isosceles right triangle with hypotenuse $2$, the area is simply $\frac{1}{2} \times \text{base} \times \text{height}$. Since $AB = AC = \sqrt{2}$, the area is $\frac{1}{2} \times \sqrt{2} \times \sqrt{2} = 1$.

JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let a triangle $A B C$ be inscribed in the circle $x^{2} - 2 (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2}$ . If the length of side $A B$ is $2$ , then the area of the $△ A B C$ is equal to:

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Given equation: $x^{2} - 2 (x + y) + y^{2} = 0$
Triangle $A B C$ is inscribed in this circle.
Given: $∠ B A C = \frac{π}{2}$ and $A B = 2$ .

Analyzing the Circle

To understand the circle's geometry, we need its center and radius.
We will convert the general equation to the standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$ .

Grouping Terms

Expand: $x^{2} - 2 x - 2 y + y^{2} = 0$
Group $x$ and $y$ terms: $(x^{2} - 2 x) + (y^{2} - 2 y) = 0$

Completing the Square

Add $(\frac{2}{2})^{2} = \frac{1}{2}$ to both sides for $x$ and $y$ .
$(x^{2} - 2 x + \frac{1}{2}) + (y^{2} - 2 y + \frac{1}{2}) = \frac{1}{2} + \frac{1}{2}$
$(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = 1$

Extracting Circle Parameters

Standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$
Center $O = (\frac{1}{2}, \frac{1}{2})$
Radius $R = 1 = 1$

The $9 0^{\circ}$ Angle Property

Given: $∠ B A C = 9 0^{\circ}$
Recall the circle theorem: An angle inscribed in a semicircle is a right angle.

Identifying the Diameter

Since $∠ B A C = 9 0^{\circ}$ , the side opposite to it must pass through the center.
Therefore, side $B C$ is the diameter of the circle.

Length of Hypotenuse $B C$

Diameter $B C = 2 \times R$
Substitute $R = 1$ :
$B C = 2 \times 1 = 2$

Applying Pythagoras Theorem

In right $△ A B C$ , we know:
Hypotenuse $B C = 2$
Side $A B = 2$ (Given)
We need side $A C$ to find the area.

Setting up Pythagoras

$A B^{2} + A C^{2} = B C^{2}$
Substitute the known values:
$(2)^{2} + A C^{2} = (2)^{2}$

Solving for Side $A C$

$2 + A C^{2} = 4$
$A C^{2} = 4 - 2 = 2$
$A C = 2$

Calculating the Area

Area of right $△ A B C = \frac{1}{2} \times base \times height$
Area $= \frac{1}{2} \times A B \times A C$
Area $= \frac{1}{2} \times 2 \times 2$
Area $= \frac{1}{2} \times 2 = 1$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We are given the equation

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

. At first glance, it is a jumble of variables, but we see the potential for a perfect circle.

Our first mission is to bring order to this chaos. By completing the square, we transform this equation into the standard form:

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = 1

Suddenly, the fog clears! We see a circle with center

(\frac{1}{2}, \frac{1}{2})

and radius

R = 1

The Geometric Insight

The

9 0^{\circ}

Beacon

Now, we turn our gaze to the triangle

A B C

. We are told

∠ B A C = 9 0^{\circ}

This is the key that unlocks the entire problem. In the world of geometry, a

9 0^{\circ}

angle inscribed in a circle is a signal—a beacon—telling us that the side opposite to it,

B C

, must be the diameter.

Since the radius is

1

, the diameter

B C

2

. We now have a right-angled triangle with hypotenuse

2

and one side

A B = 2

The Final Calculation

Pythagoras and Elegance

Using the Pythagorean theorem,

A B^{2} + A C^{2} = B C^{2}

, we find:

A C^{2} = 4 - 2 = 2

This implies

A C = 2

The area of a right-angled triangle is given by the formula

\frac{1}{2} \times base \times height

. Substituting our values, we get:

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

A clean, elegant result. Remember, in JEE, the most complex problems often yield to the most fundamental principles. The final answer is 1.

Animated Solution for Mathematics - Circles: Let a triangle $A B C$ be inscribed in the circle $x^{2} - 2 (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2}$ . If the length of side $A B$ is $2$ , then the area of the $△ A B C$ is equal to:

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Analyzing the Circle

Grouping Terms

Completing the Square

Extracting Circle Parameters

The $9 0^{\circ}$ Angle Property

Identifying the Diameter

Length of Hypotenuse $B C$

Applying Pythagoras Theorem

Setting up Pythagoras

Solving for Side $A C$

Calculating the Area

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Insight

The Final Calculation

Similar Questions

If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

In the circle given below, let $O A = 1$ unit, $O B = 13$ unit and $P Q ∥ O B$ . Then, the area of the triangle PQB (in square units) is:

Let $A B C$ be the triangle with $A B = 1, A C = 3$ and $∠ B A C = \frac{π}{2}$ . If a circle of radius $r > 0$ touches the sides $A B, A C$ and also touches internally the circumcircle of the triangle $A B C$ , then the value of $r$ is _____________.

Let the line $x + y = 1$ meet the circle $x^{2} + y^{2} = 4$ at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to:

Let the parabola $y = x^{2} + p x - 3$ , meet the coordinate axes at the points P, Q and R. If the circle C with centre at $(- 1, - 1)$ passes through the points P, Q and R, then the area of $△ P QR$ is:

The circle $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is $x + y - x y + k (x^{2} + y^{2})^{1/2} = 0$ . Find $k$ .

A rectangle $R$ with end points of the one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x - y + 4 = 0$ , then the area of $R$ is _______.

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Analyzing the Circle

Grouping Terms

Completing the Square

Extracting Circle Parameters

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 90∘ Angle Property

Identifying the Diameter

Length of Hypotenuse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Applying Pythagoras Theorem

Setting up Pythagoras

Solving for Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Calculating the Area

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Insight

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273​ sq. units then c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the circle given below, let OA=1 unit, OB=13 unit and PQ∥OB. Then, the area of the triangle PQB (in square units) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

The $9 0^{\circ}$ Angle Property

Length of Hypotenuse $B C$

Solving for Side $A C$

If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

In the circle given below, let $O A = 1$ unit, $O B = 13$ unit and $P Q ∥ O B$ . Then, the area of the triangle PQB (in square units) is:

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is