Let $C$ be a circle which passes through $(-2,1)$ and $(0,3)$ and whose centre lies on the line $y-x...

The general equation of a circle is:$ (x-h)^2 + (y-k)^2 = r^2 $Here, $(h, k)$ represents the center of the circle, $r$ is its radius, and $(x, y)$ is any point on the circle.Given that the circle's center lies on the line $y = x + 1$, we have $k = h + 1$.We can substitute this relationship and use the following two points through which the circle passes: $(-2, 1)$ and $(0, 3)$.By setting the distances equal, we obtain:$ (-2-h)^2 + (1-(h+1))^2 = (0-h)^2 + (3-(h+1))^2 $Simplifying, we have:$ (2+h)^2 + h^2 = h^2 + (2-h)^2 $This simplifies to:$ (2+h)^2 = (2-h)^2 $Solving for $h$, we find:$ h = 0, \quad k = 1 $The circle's center is $(0, 1)$ with radius $2$ (the same as the distance from $(0, 1)$ to $(0, 3)$).Next, we verify which given points lie on this circle:Observing the point $(\sqrt{2}+1, 1)$, it results in an irrational distance, hence it can be dismissed. Similarly, $(1, \sqrt{3}-1)$ results in an irrational value, ruling it out.Checking the remaining options:$ \begin{align*} (1, \sqrt{2}+1): &\quad 1^2 + (\sqrt{2} + 1 - 1)^2 = 1 + (\sqrt{2})^2 = 1 + 2 = 3 \quad (\neq 4) \\ (1, \sqrt{3}+1): &\quad 1^2 + (\sqrt{3} + 1 - 1)^2 = 1 + (\sqrt{3})^2 = 1 + 3 = 4 \end{align*} $Thus, the point $(1, \sqrt{3}+1)$ lies on the circle.

Consider the tangent lines to the circle $x^2+y^2=1$ at points $P=(1,0)$ and $Q=\left(\frac{1}{\sqrt...

$$ \begin{aligned} &\begin{gathered} \tan (x)=\frac{1 / \sqrt{2}}{1 / \sqrt{2}}=1 \\ \Rightarrow x=45^{\circ} \end{gathered}\\ &\text { Now, you can simply use the sum of angles in quadrilaterals. }\\ &x+\theta+90+90=360 \Rightarrow \theta=135^{\circ} \end{aligned} $$

What is the locus of the center of circles passing through the origin $(0,0)$ with fixed radius?...

To determine the locus of the centers of circles passing through the origin $(0,0)$ with a fixed radius, let's analyze the general equation of a circle:$ (x-h)^2 + (y-k)^2 = r^2 $Here, $(h, k)$ represents the center of the circle, and $r$ is the radius.Given that the circle passes through the origin $(0,0)$, we substitute $x = 0$ and $y = 0$ into the circle's equation:$ (0-h)^2 + (0-k)^2 = r^2 $This simplifies to:$ h^2 + k^2 = r^2 $The equation $h^2 + k^2 = r^2$ describes a circle with a radius of $r$ and centered at the origin $(0,0)$. Hence, the locus of the centers of such circles is a circle itself, centered at the origin with the same fixed radius $r$.

Circle - IAT-IISER Maths PYQs

The general equation of a circle is:

(x-h)%5E2%20%2B%20(y-k)%5E2%20%3D%20r%5E2

Here,

(h%2C%20k)

represents the center of the circle,

r

is its radius, and

(x%2C%20y)

is any point on the circle.Given that the circle's center lies on the line

y%20%3D%20x%20%2B%201

, we have

k%20%3D%20h%20%2B%201

.We can substitute this relationship and use the following two points through which the circle passes:

(-2%2C%201)

and

(0%2C%203)

.By setting the distances equal, we obtain:

(-2-h)%5E2%20%2B%20(1-(h%2B1))%5E2%20%3D%20(0-h)%5E2%20%2B%20(3-(h%2B1))%5E2

Simplifying, we have:

(2%2Bh)%5E2%20%2B%20h%5E2%20%3D%20h%5E2%20%2B%20(2-h)%5E2

This simplifies to:

(2%2Bh)%5E2%20%3D%20(2-h)%5E2

Solving for

h

, we find:

h%20%3D%200%2C%20%5Cquad%20k%20%3D%201

The circle's center is

(0%2C%201)

with radius

2

(the same as the distance from

(0%2C%201)

(0%2C%203)

).Next, we verify which given points lie on this circle:Observing the point

(%5Csqrt%7B2%7D%2B1%2C%201)

, it results in an irrational distance, hence it can be dismissed. Similarly,

(1%2C%20%5Csqrt%7B3%7D-1)

results in an irrational value, ruling it out.Checking the remaining options:

%5Cbegin%7Balign*%7D%20(1%2C%20%5Csqrt%7B2%7D%2B1)%3A%20%26%5Cquad%201%5E2%20%2B%20(%5Csqrt%7B2%7D%20%2B%201%20-%201)%5E2%20%3D%201%20%2B%20(%5Csqrt%7B2%7D)%5E2%20%3D%201%20%2B%202%20%3D%203%20%5Cquad%20(%5Cneq%204)%20%5C%5C%20(1%2C%20%5Csqrt%7B3%7D%2B1)%3A%20%26%5Cquad%201%5E2%20%2B%20(%5Csqrt%7B3%7D%20%2B%201%20-%201)%5E2%20%3D%201%20%2B%20(%5Csqrt%7B3%7D)%5E2%20%3D%201%20%2B%203%20%3D%204%20%5Cend%7Balign*%7D

Thus, the point

(1%2C%20%5Csqrt%7B3%7D%2B1)

lies on the circle.

To determine the locus of the centers of circles passing through the origin

(0%2C0)

with a fixed radius, let's analyze the general equation of a circle:

(x-h)%5E2%20%2B%20(y-k)%5E2%20%3D%20r%5E2

Here,

(h%2C%20k)

represents the center of the circle, and

r

is the radius.Given that the circle passes through the origin

(0%2C0)

, we substitute

x%20%3D%200

and

y%20%3D%200

into the circle's equation:

(0-h)%5E2%20%2B%20(0-k)%5E2%20%3D%20r%5E2

This simplifies to:

h%5E2%20%2B%20k%5E2%20%3D%20r%5E2

The equation

h%5E2%20%2B%20k%5E2%20%3D%20r%5E2

describes a circle with a radius of

r

and centered at the origin

(0%2C0)

. Hence, the locus of the centers of such circles is a circle itself, centered at the origin with the same fixed radius

r

Circle

High-Yield Trend

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3 MCQs

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Circle

High-Yield Trend

Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Questions
3 MCQs