Let P(a₁, b₁) and Q(a₂, b₂) be two distinct points on a circle with center C(√2, √3). Let O be the o

Q: Wait, why is $\triangle OCP$ a right-angled triangle? The problem says $OC \perp CP$, but does that mean the angle at $C$ is $90^\circ$?

Yes, exactly! The notation $OC \perp CP$ literally means the line segment $OC$ is perpendicular to the line segment $CP$. In geometry, perpendicular lines meet at a $90^\circ$ angle. Therefore, the triangle formed by vertices $O$, $C$, and $P$ must have a right angle at vertex $C$.

Q: Why can't we just write the equation of the circle and substitute the points?

You absolutely could, but that is the 'brute force' path. It would involve solving for the radius $r$ and then dealing with the coordinates $(a_1, b_1)$ algebraically, which is prone to calculation errors. The 'elegant' path uses the geometric properties of the triangle to find the squared distances directly, which is exactly what the question asks for.

Q: Does the fact that $P$ and $Q$ are distinct matter?

In this specific problem, the distinctness ensures that we have two separate points, but the geometric relationship holds for any point on the circle satisfying the perpendicularity condition. It confirms that the logic applies to both $P$ and $Q$ independently, allowing us to sum their squared coordinates.

Question

Let $P(a_1, b_1)$ and $Q(a_2, b_2)$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then $a_1^2 + a_2^2 + b_1^2 + b_2^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today we tackle a problem that seems to be about coordinates, but is truly about the elegance of right-angled triangles hidden in plain sight.

When you see a problem involving a circle, an origin, and perpendicularity, do not rush to write the equation of the circle. Instead, pause and visualize.

Imagine the origin $O$ at $(0,0)$ and the center $C$ at $(\sqrt{2}, \sqrt{3})$. The problem states that $OC \perp CP$. This is not just a line; it is a gateway to the Pythagorean theorem.

We calculate the distance $OC$ using the distance formula:

$$OC = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{3} - 0)^2} = \sqrt{2 + 3} = \sqrt{5}$$

This value serves as our base.

### The Hidden Radius

The area of $	riangle OCP$ is given as $\frac{\sqrt{35}}{2}$. Since it is a right-angled triangle at $C$, the area is simply $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, which is $\frac{1}{2} 	imes OC 	imes CP$.

Substituting our known values, we get:

$$\frac{1}{2} 	imes \sqrt{5} 	imes CP = \frac{\sqrt{35}}{2}$$

The $\frac{1}{2}$ cancels out, leaving $\sqrt{5} 	imes CP = \sqrt{35}$. Dividing both sides by $\sqrt{5}$, we find that $CP = \sqrt{7}$.

This $CP$ is the radius of our circle. It represents the constant distance from the center $C$ to any point on the circumference.

### The Pythagorean Insight

Now, consider $	riangle OCP$ again. It is right-angled at $C$.

Thus, by the Pythagorean theorem, the square of the hypotenuse $OP$ is the sum of the squares of the legs:

$$OP^2 = OC^2 + CP^2$$

Substituting our values:

$$OP^2 = (\sqrt{5})^2 + (\sqrt{7})^2 = 5 + 7 = 12$$

Since $P$ is $(a_1, b_1)$, the square of the distance from the origin is $OP^2 = a_1^2 + b_1^2 = 12$. By symmetry, the exact same logic applies to point $Q(a_2, b_2)$.

Thus, $OQ^2 = a_2^2 + b_2^2 = 12$.

### The Final Summation

We are asked to find $a_1^2 + a_2^2 + b_1^2 + b_2^2$. Rearranging these terms, we get $(a_1^2 + b_1^2) + (a_2^2 + b_2^2)$.

We have already calculated both groups to be $12$. Therefore, the sum is:

$$12 + 12 = 24$$

This problem teaches us that the most complex-looking coordinate geometry questions often collapse into simple, beautiful triangles if we just look for the right angles. The final answer is **24**.

Enter Numerical Value:

Visualized Solution

Setting up the Coordinate System

The Circle and Points $P, Q$

The Perpendicularity Condition

Calculating Distance $O C$

Area of $△ O C P$

Setting up the Area Equation

Calculating the Radius $C P$

Applying Pythagoras Theorem

Calculating $O P^{2}$

Symmetry for Triangle $O C Q$

Calculating $O Q^{2}$

The Final Summation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Radius

The Pythagorean Insight

The Final Summation

Similar Questions

Let $P$ and $Q$ be two distinct points on a circle which has center at $C (2, 3)$ and which passes through origin $O$ . If $O C$ is perpendicular to both the line segments $C P$ and $C Q$ , then the set ${P, Q}$ 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:

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________

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:

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 :

Let a circle of radius 4 pass through the origin $O$ , the points $A (- 3 a, 0)$ and $B (0, - 2 b)$ , where $a$ and $b$ are real parameters and $ab \neq = 0$ . Then the locus of the centroid of $△ O A B$ is a circle of radius

The triangle $P QR$ is inscribed in the circle $x^{2} + y^{2} = 25$ . If $Q$ and $R$ have co-ordinates $(3, 4)$ and $(- 4, 3)$ respectively, then $∠ QP R$ is equal to

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

The abscissa of the two points $A$ and $B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and their ordinates are the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Find the equation and the radius of the circle with $A B$ as diameter.

Enter Numerical Value:

Visualized Solution

Setting up the Coordinate System

The Circle and Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P,Q

The Perpendicularity Condition

Calculating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OC

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OCP

Setting up the Area Equation

Calculating the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CP

Applying Pythagoras Theorem

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OP2

Symmetry for Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OCQ

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ2

The Final Summation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Radius

The Pythagorean Insight

The Final Summation

Similar Questions

.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; } If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3​)x+2(4−6​)y=k+63​+86​,k>0, touch internally at the point P(α,β), then (α+3​)2+(β+6​)2 is equal to ________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a triangle ABC be inscribed in the circle x2−2​(x+y)+y2=0 such that ∠BAC=2π​. If the length of side AB is 2​, then the area of the △ABC is equal to:

.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; } The triangle PQR is inscribed in the circle x2+y2=25. If Q and R have co-ordinates (3,4) and (−4,3) respectively, then ∠QPR is equal to

.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 Circle and Points $P, Q$

Calculating Distance $O C$

Area of $△ O C P$

Calculating the Radius $C P$

Calculating $O P^{2}$

Symmetry for Triangle $O C Q$

Calculating $O Q^{2}$

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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:

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 :

The triangle $P QR$ is inscribed in the circle $x^{2} + y^{2} = 25$ . If $Q$ and $R$ have co-ordinates $(3, 4)$ and $(- 4, 3)$ respectively, then $∠ QP R$ is equal to

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