Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

(x - h)^2 + (y - k)^2 = r^2

That's it. That's the standard form of the equation of a circle.

(x - h)^2 + (y - k)^2 = r^2

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

(x + 1)^2 + (y - 4)^2 = 9

That's the standard form. From this, you can immediately read off the centre $(-1, 4)$ and radius $3$ .

Why This Form Matters

The standard form is the most useful because it gives you the centre and radius at a glance. In exams, you'll often be given an expanded form like $x^2 + y^2 - 6x + 4y - 12 = 0$ and asked to rewrite it in standard form by completing the square — that's the next step in your learning, but the standard form itself is the destination.

For now: centre tells you where, radius tells you how big, and the equation tells you which points belong.

The Standard Form of a Circle's Equation is one of the first results in the NCERT Class 11 Mathematics chapter on Conic Sections, matching searches like "equation of a circle: definition, formula and examples" or "conic sections important questions class 11 maths". Recognising centre and radius directly from this form is also a routine, quick-scoring question type in CBSE boards, JEE Main, and state CET coordinate geometry sections.

A circle touching both axes in the first quadrant with radius $a$ has its center at $(a, a)$ , leading to the equation $(x-a)^2 + (y-a)^2 = a^2$ .

To find the equation of a circle, we primarily need two pieces of information: its center and its radius. The standard form of a circle's equation directly uses these values.

(h, k)

Here, $h$ and $k$ are the x and y coordinates of the center, respectively, and $r$ is the radius. Our task is to use the given conditions to determine $h$ , $k$ , and $r$ .

The problem states two crucial conditions:

The circle touches both the x-axis and the y-axis.
It is located in the first quadrant.
Its radius is $a$ .

Let's break down how these conditions help us find the center $(h, k)$ and radius $r$ .

Identify the radius:

The problem explicitly states that the radius of the circle is $a$ .

So, we have $r = a$ .
Determine the coordinates of the center $(h, k)$ :
- Touching the x-axis: If a circle touches the x-axis, the perpendicular distance from its center to the x-axis must be equal to its radius. This distance is simply the absolute value of the y-coordinate of the center. Since the circle is in the first quadrant, its center's y-coordinate must be positive. Therefore, $k = r$ .
- Touching the y-axis: Similarly, if a circle touches the y-axis, the perpendicular distance from its center to the y-axis must be equal to its radius. This distance is the absolute value of the x-coordinate of the center. Since the circle is in the first quadrant, its center's x-coordinate must also be positive. Therefore, $h = r$ .
Combining these, and knowing $r=a$ , the center of the circle must be $(a, a)$ .
Tip
For a circle touching both axes, its center will always be $(\pm r, \pm r)$ , where the signs depend on the quadrant.
- First Quadrant: $(r, r)$
- Second Quadrant: $(-r, r)$
- Third Quadrant: $(-r, -r)$
- Fourth Quadrant: $(r, -r)$
Substitute the center and radius into the standard equation:

Now we have all the necessary components:
- Center $(h, k) = (a, a)$
- Radius $r = a$
Substitute these values into the standard equation $(x-h)^2 + (y-k)^2 = r^2$ :

(x-a)^2 + (y-a)^2 = a^2

This is the equation of the circle. We can also expand it to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$:

x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2

x^2 + y^2 - 2ax - 2ay + a^2 = 0

Both forms are correct, but the first one directly reflects the center and radius.

✓Final answer

\boxed{(x-a)^2 + (y-a)^2 = a^2}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-M1 markMCQ
$2 x-3 y+12=0$
›Reveal solutionSolution
The circle's center is the intersection of its two given diameters, and its radius comes from the given area. This gives $x^2+y^2+6x-4y-36=0$ , matching option (D).

Step-by-step reasoning

Find the centre (intersection of the diameters).

$2x-3y+12=0, \qquad x+4y-5=0$
From the second: $x=5-4y$ . Substituting into the first:
$2(5-4y)-3y+12=0 \Rightarrow 10-8y-3y+12=0 \Rightarrow 22-11y=0 \Rightarrow y=2$
Then $x=5-4(2)=-3$ . Centre $=(-3,2)$ .

Find the radius from the area.

$\pi r^2 = 154 \quad (\pi\approx\tfrac{22}{7}) \quad\Rightarrow\quad r^2 = 154\times\frac{7}{22}=49 \quad\Rightarrow\quad r=7$

Write and expand the circle equation.

$(x+3)^2+(y-2)^2=49$ $x^2+6x+9+y^2-4y+4=49 \quad\Rightarrow\quad x^2+y^2+6x-4y-36=0$
Watch out

The centre is the intersection of the two diameters, not just any point on either line.

Tip

Verify by substituting the centre into the final equation: it should give $-r^2$ . At $(-3,2)$ : $9+4-18-8-36=-49=-r^2$ ✓.

✓Final answer
$The correct option is (D) .$
KCET 2021Set A-11 markMCQ
$y = \alpha x^2 - 6x + \beta$
›Reveal solutionSolution
The parabola passes through $(0,2)$ and has a horizontal tangent at $x = \frac{3}{2}$ . Using the point condition gives $\beta = 2$ , and the derivative condition gives $\alpha = 2$ . The correct pair is $\alpha = 2, \beta = 2$ , option (C).

The key idea here is that a tangent parallel to the $x$ -axis means the slope of the tangent is zero. For a curve $y = f(x)$ , the slope of the tangent at any point is given by $\frac{dy}{dx}$ . So "tangent parallel to $x$ -axis" translates directly to $\frac{dy}{dx} = 0$ at that $x$ -coordinate.

We also have a point that lies on the parabola. Substituting that point into the equation gives a direct relation between $\alpha$ and $\beta$ .

Let’s work through it step by step.

Use the point $(0,2)$ . The parabola is $y = \alpha x^2 - 6x + \beta$ . Substituting $x = 0$ , $y = 2$ :

$2 = \alpha(0)^2 - 6(0) + \beta \quad\Rightarrow\quad \beta = 2.$

Find the derivative. Differentiate $y$ with respect to $x$ :

$\frac{dy}{dx} = 2\alpha x - 6.$

Apply the tangent condition. At $x = \frac{3}{2}$ , the tangent is parallel to the $x$ -axis, so $\frac{dy}{dx} = 0$ :

$2\alpha \left(\frac{3}{2}\right) - 6 = 0 \quad\Rightarrow\quad 3\alpha - 6 = 0 \quad\Rightarrow\quad \alpha = 2.$

Check the options. We have $\alpha = 2$ , $\beta = 2$ . This matches option (C).

Watch out

A common mistake is to forget that "parallel to $x$ -axis" means slope zero, not just "horizontal" in the sense of a constant function. Also, be careful: the derivative is set to zero at the given $x$ , not at the given point $(0,2)$ .

Tip

You could also think of this geometrically: a parabola $y = ax^2 + bx + c$ has its vertex where the tangent is horizontal. Here, the vertex occurs at $x = \frac{3}{2}$ , so you could directly use the vertex formula $x = -\frac{b}{2a}$ with $b = -6$ to get $-\frac{-6}{2\alpha} = \frac{3}{2}$ , which gives $\alpha = 2$ — same result, faster.

✓Final answer
$\alpha = 2$
COMEDK 2021Set 20211 markMCQ
${x^2} + {y^2} - 2x - 4y + 1 = 0$
›Reveal solutionSolution

Check: centre = (1, 2), r = sqrt(1 + 4 - 1) = sqrt(4) = 2. Distance from centre to the X-axis = 2 = r, so it touches the X-axis. Correct.

Concept: A circle that touches the X-axis has radius equal to the absolute value of the y-coordinate of its centre (the perpendicular distance from the centre to the line y = 0).

Centre (h, k) = (1, 2) -> radius r = |k| = 2.

Equation: (x - 1)^2 + (y - 2)^2 = 2^2

x^2 - 2x + 1 + y^2 - 4y + 4 = 4

x^2 + y^2 - 2x - 4y + 1 = 0

Check: centre = (1, 2), r = sqrt(1 + 4 - 1) = sqrt(4) = 2. Distance from centre to the X-axis = 2 = r, so it touches the X-axis. Correct.

✓Final answer
${x^2} + {y^2} - 2x - 4y + 1 = 0$

ANSWER: A
COMEDK 2021Set 20211 markMCQ
$2{x^2} + 2{y^2} + 3x + 4y + {9 \over 8} = 0$
›Reveal solutionSolution

The four options are single numbers, so they refer to the radius; the radius is 1.

Concept: general equation of a circle x^2 + y^2 + 2gx + 2fy + c = 0 has centre (-g, -f) and radius sqrt(g^2 + f^2 - c).

First make the coefficients of x^2 and y^2 equal to 1 by dividing throughout by 2:

2x^2 + 2y^2 + 3x + 4y + 9/8 = 0

=> x^2 + y^2 + (3/2)x + 2y + 9/16 = 0.

So 2g = 3/2 => g = 3/4; 2f = 2 => f = 1; c = 9/16.

Centre = (-3/4, -1).

Radius = sqrt(g^2 + f^2 - c) = sqrt(9/16 + 1 - 9/16) = sqrt(1) = 1.

The four options are single numbers, so they refer to the radius; the radius is 1.

✓Final answer
$The correct option is (A) — 1$

ANSWER: A
COMEDK 2021Set 20211 markMCQ
${x^2} + {y^2} - 2x - 4y - 20 = 0$
›Reveal solutionSolution

Equation: (x - 1)^2 + (y - 2)^2 = 25 => x^2 - 2x + 1 + y^2 - 4y + 4 = 25 => x^2 + y^2 - 2x - 4y + 5 - 25 = 0 => x^2 + y^2 - 2x - 4y - 20 = 0.

Concept: circle with centre (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2, where r is the distance from the centre to any point on the circle.

Centre (1, 2), point on circle (4, 6).

r = sqrt((4 - 1)^2 + (6 - 2)^2) = sqrt(9 + 16) = sqrt(25) = 5.

Equation: (x - 1)^2 + (y - 2)^2 = 25

=> x^2 - 2x + 1 + y^2 - 4y + 4 = 25

=> x^2 + y^2 - 2x - 4y + 5 - 25 = 0

=> x^2 + y^2 - 2x - 4y - 20 = 0.

✓Final answer
${x^2} + {y^2} - 2x - 4y - 20 = 0$

ANSWER: A
KCET 2018Set A-11 markMCQ
$3x - 4y + 2 = 0$
›Reveal solutionSolution

Parallel lines have equal slopes, so keep the $x$ - and $y$ -coefficients unchanged and fix the constant by making the line pass through the given point.

Step 1 — Use the parallelism condition.

The slope of $3x - 4y + 2 = 0$ is
$m = -\frac{\text{coefficient of }x}{\text{coefficient of }y} = -\frac{3}{-4} = \frac{3}{4}$
Any line parallel to it has the same slope, so it can be written by changing only the constant term:
$3x - 4y + c = 0$
This is the standard "family of parallel lines" trick — it saves converting to slope-intercept form.

Step 2 — Impose the point condition.

The line must pass through $(-2, 3)$ , so those coordinates must satisfy it:
$3(-2) - 4(3) + c = 0$ $-6 - 12 + c = 0 \;\Rightarrow\; c = 18$
Step 3 — Write the equation.
$3x - 4y + 18 = 0$
Step 4 — Verify.

Substitute $(-2,3)$ : $\;3(-2) - 4(3) + 18 = -6 - 12 + 18 = 0 \ \checkmark$ . And its slope is $3/4$ , matching the given line. ✓

Options (C) and (D) have $+4y$ , giving slope $-3/4$ — those are lines perpendicular-ish in coefficient pattern, certainly not parallel. Option (B) has the right direction but fails the point test: $-6 - 12 - 18 = -36 \neq 0$ .

✓Final answer
$3x - 4y + 18 = 0$

ANSWER: A

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Q.Find the equation of the circle which touches the both axes in first quadrant and whose radius is aaa.

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Q.Find the equation of the circle which touches the both axes in first quadrant and whose radius is $a$ .

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