Question 1

The circle x² + y² − 4x + 2y − 4 = 0 intersects the x-axis at two points. The distance between these two points is:

Accepted Answer

To find the intersection with the x-axis, set y = 0 in the circle equation:

x² + 0² − 4x + 2(0) − 4 = 0
x² − 4x − 4 = 0

Using the quadratic formula:
x = (4 ± √(16 + 16)) / 2
x = (4 ± √32) / 2
x = (4 ± 4√2) / 2
x = 2 ± 2√2

The two intersection points are (2 − 2√2, 0) and (2 + 2√2, 0).

The distance between them is:
|(2 + 2√2) − (2 − 2√2)| = |4√2| = 4√2

Alternatively, for a circle x² + y² + 2gx + 2fy + c = 0 intersecting the x-axis, the distance between intersection points is 2√(g² − c).
Here, 2g = −4 → g = −2, c = −4.
Distance = 2√((−2)² − (−4)) = 2√(4 + 4) = 2√8 = 4√2

Question 2

The equation of the tangent to the circle x² + y² = 25 at the point (3, 4) is:

Accepted Answer

For a circle x² + y² = r², the equation of the tangent at a point (x₁, y₁) on the circle is:
xx₁ + yy₁ = r²

Here, r² = 25, and the point is (3, 4).

The tangent equation is:
x(3) + y(4) = 25
3x + 4y = 25

Verification: The point (3, 4) lies on the circle since 3² + 4² = 9 + 16 = 25. ✓
The tangent at (3, 4) is perpendicular to the radius at that point.
Radius slope = 4/3, so tangent slope = −3/4.
Using point-slope form: y − 4 = (−3/4)(x − 3)
4(y − 4) = −3(x − 3)
4y − 16 = −3x + 9
3x + 4y = 25 ✓

Question 3

The equation of the circle passing through the origin with centre at (3, 4) is:

Accepted Answer

The general equation of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r².

Given: Centre = (3, 4), and the circle passes through the origin (0, 0).

Step 1: Find the radius using the distance formula from centre to origin.
r = √[(0 − 3)² + (0 − 4)²] = √[9 + 16] = √25 = 5

Step 2: Write the equation of the circle.
(x − 3)² + (y − 4)² = 5² = 25

Step 3: Expand to verify.
x² − 6x + 9 + y² − 8y + 16 = 25
x² + y² − 6x − 8y + 25 − 25 = 0
x² + y² − 6x − 8y = 0

Verification: At origin (0, 0): 0 + 0 − 0 − 0 = 0 ✓

Question 4

The radius of the circle x² + y² − 4x + 6y − 3 = 0 is:

Accepted Answer

The general form of a circle is: x² + y² + 2gx + 2fy + c = 0

Given equation: x² + y² − 4x + 6y − 3 = 0

Step 1: Identify coefficients by comparing with general form.
2g = −4  ⟹  g = −2
2f = 6   ⟹  f = 3
c = −3

Step 2: Apply the radius formula.
For a circle x² + y² + 2gx + 2fy + c = 0, the radius r = √(g² + f² − c)

Step 3: Calculate.
r = √[(−2)² + (3)² − (−3)]
r = √[4 + 9 + 3]
r = √16
r = 4

Question 5

The centre and radius of the circle (x + 2)² + (y − 3)² = 49 are respectively:

Accepted Answer

The standard form of a circle is: (x − h)² + (y − k)² = r²
where (h, k) is the centre and r is the radius.

Given equation: (x + 2)² + (y − 3)² = 49

Step 1: Rewrite in standard form.
(x − (−2))² + (y − 3)² = 49

Step 2: Identify the centre and radius.
Centre: (h, k) = (−2, 3)
Radius: r = √49 = 7

Step 3: Verify by expanding.
(x + 2)² + (y − 3)² = 49
x² + 4x + 4 + y² − 6y + 9 = 49
x² + y² + 4x − 6y + 13 − 49 = 0
x² + y² + 4x − 6y − 36 = 0

Using general form formula: Centre = (−g, −f) = (−2, 3) ✓, Radius = √(4 + 9 + 36) = √49 = 7 ✓

Question 6

If the circle x² + y² + 2gx + 2fy + c = 0 passes through the origin, then:

Accepted Answer

The general equation of a circle is: x² + y² + 2gx + 2fy + c = 0

Given: The circle passes through the origin (0, 0).

Step 1: Substitute the point (0, 0) into the circle equation.
(0)² + (0)² + 2g(0) + 2f(0) + c = 0
0 + 0 + 0 + 0 + c = 0
c = 0

Step 2: Conclusion.
If a circle passes through the origin, then the constant term c must equal 0.

Step 3: Verify with an example.
Circle x² + y² − 4x − 6y = 0 (c = 0) passes through (0, 0): 0 + 0 − 0 − 0 = 0 ✓
Circle x² + y² − 4x − 6y + 5 = 0 (c = 5) does not pass through (0, 0): 0 + 0 − 0 − 0 + 5 = 5 ≠ 0 ✓

Question 7

The general form of a circle is x² + y² − 8x + 6y + 9 = 0. What is the radius of this circle?

Accepted Answer

The general form of a circle is x² + y² + 2gx + 2fy + c = 0.

Comparing with x² + y² − 8x + 6y + 9 = 0:
2g = −8 → g = −4
2f = 6 → f = 3
c = 9

The centre is (−g, −f) = (4, −3).

The radius is given by: r = √(g² + f² − c)
r = √((−4)² + 3² − 9)
r = √(16 + 9 − 9)
r = √16
r = 4

The radius of the circle is 4.

Question 8

A circle passes through the points (1, 0) and (3, 0). If the centre lies on the line y = x, what is the radius of the circle?

Accepted Answer

Since the circle passes through both (1, 0) and (3, 0), the centre must be equidistant from these two points.

Let the centre be (a, a) since it lies on y = x.

Distance from (a, a) to (1, 0):
d₁ = √((a − 1)² + (a − 0)²) = √((a − 1)² + a²)

Distance from (a, a) to (3, 0):
d₂ = √((a − 3)² + (a − 0)²) = √((a − 3)² + a²)

Since both points lie on the circle: d₁ = d₂
√((a − 1)² + a²) = √((a − 3)² + a²)

Squaring both sides:
(a − 1)² + a² = (a − 3)² + a²
(a − 1)² = (a − 3)²
a² − 2a + 1 = a² − 6a + 9
−2a + 1 = −6a + 9
4a = 8
a = 2

So the centre is (2, 2).

Radius = distance from (2, 2) to (1, 0):
r = √((2 − 1)² + (2 − 0)²) = √(1 + 4) = √5

The radius is √5.

Question 9

The circle x² + y² = 16 is intersected by the line y = 2. At how many points do they intersect?

Accepted Answer

The circle is x² + y² = 16 with centre (0, 0) and radius 4.

Substitute y = 2 into the circle equation:
x² + 2² = 16
x² + 4 = 16
x² = 12
x = ±2√3

Since x² = 12 has two real solutions (x = 2√3 and x = −2√3), the line intersects the circle at two distinct points: (2√3, 2) and (−2√3, 2).

Alternatively, check the distance from the centre (0, 0) to the line y = 2:
Distance d = |2 − 0| = 2

Since d = 2 < r = 4 (the distance is less than the radius), the line intersects the circle at exactly 2 points.

Question 10

The equation of the circle with centre (3, −2) and radius 5 is:

Accepted Answer

The standard form of a circle with centre (h, k) and radius r is: (x − h)² + (y − k)² = r².

Given: centre = (3, −2), radius = 5.

Substituting into the standard form:
(x − 3)² + (y − (−2))² = 5²
(x − 3)² + (y + 2)² = 25

Expanding to verify:
x² − 6x + 9 + y² + 4y + 4 = 25
x² + y² − 6x + 4y + 13 = 25
x² + y² − 6x + 4y − 12 = 0

The correct equation is (x − 3)² + (y + 2)² = 25.

Question 11

The general form of a circle is x² + y² − 8x + 6y + 9 = 0. The radius of this circle is:

Accepted Answer

The general form of a circle is x² + y² + 2gx + 2fy + c = 0.

Comparing with x² + y² − 8x + 6y + 9 = 0:
2g = −8 → g = −4
2f = 6 → f = 3
c = 9

The centre is (−g, −f) = (4, −3).

The radius is given by: r = √(g² + f² − c)
r = √((−4)² + 3² − 9)
r = √(16 + 9 − 9)
r = √16
r = 4

The radius is 4.

Question 12

A circle passes through the points (1, 0) and (−1, 0), and its centre lies on the line y = x. The equation of the circle is:

Accepted Answer

Since the circle passes through (1, 0) and (−1, 0), the centre must lie on the perpendicular bisector of the chord joining these two points.

The midpoint of (1, 0) and (−1, 0) is (0, 0).
The chord is horizontal (along the x-axis), so the perpendicular bisector is vertical: x = 0 (the y-axis).

The centre lies on both x = 0 and y = x.
Substituting x = 0 into y = x: y = 0.
So the centre is (0, 0).

The radius is the distance from (0, 0) to (1, 0):
r = √((1 − 0)² + (0 − 0)²) = √1 = 1

The equation is: x² + y² = 1

Question 13

A circle passes through the point (2, 3) and has its centre at (−1, 1). Find the equation of the circle in the form x² + y² + 2gx + 2fy + c = 0.

Accepted Answer

Step 1: Find the radius using the distance formula between centre (−1, 1) and point (2, 3).
r = √[(2−(−1))² + (3−1)²] = √[9 + 4] = √13.

Step 2: Write the standard form with centre (−1, 1) and radius √13:
(x + 1)² + (y − 1)² = 13.

Step 3: Expand:
x² + 2x + 1 + y² − 2y + 1 = 13
x² + y² + 2x − 2y + 2 − 13 = 0
x² + y² + 2x − 2y − 11 = 0.

Step 4: Convert to the form x² + y² + 2gx + 2fy + c = 0:
Comparing: 2g = 2 → g = 1; 2f = −2 → f = −1; c = −11.
Answer: x² + y² + 2x − 2y − 11 = 0.

Question 14

The circle x² + y² − 6x + 4y − 12 = 0 is touched by a line at the point (7, 2). Find the equation of the tangent line.

Accepted Answer

Step 1: Convert the circle equation to standard form by completing the square.
x² − 6x + y² + 4y = 12
(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x − 3)² + (y + 2)² = 25.
Centre: C = (3, −2), Radius: r = 5.

Step 2: Verify that (7, 2) lies on the circle.
(7 − 3)² + (2 + 2)² = 16 + 16 = 32 ≠ 25.
This indicates an error in the problem statement. However, proceeding with the tangent formula:

Step 3: The tangent at point (x₁, y₁) on circle x² + y² + 2gx + 2fy + c = 0 is:
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
Here: g = −3, f = 2, c = −12, and (x₁, y₁) = (7, 2).

Step 4: Substitute:
7x + 2y − 3(x + 7) + 2(y + 2) − 12 = 0
7x + 2y − 3x − 21 + 2y + 4 − 12 = 0
4x + 4y − 29 = 0.

Alternatively, using the slope method:
Slope of radius from (3, −2) to (7, 2): m_r = (2 − (−2))/(7 − 3) = 4/4 = 1.
Slope of tangent: m_t = −1/1 = −1.
Tangent equation: y − 2 = −1(x − 7) → y − 2 = −x + 7 → x + y − 9 = 0.

Question 15

Two circles have equations x² + y² = 9 and (x − 4)² + (y − 3)² = 16. Find the length of the common chord (if it exists).

Accepted Answer

Step 1: Identify the circles.
Circle 1: Centre C₁ = (0, 0), Radius r₁ = 3.
Circle 2: Centre C₂ = (4, 3), Radius r₂ = 4.

Step 2: Find the distance between centres.
d = √[(4 − 0)² + (3 − 0)²] = √[16 + 9] = √25 = 5.

Step 3: Check if circles intersect.
For two circles to intersect at two points: |r₁ − r₂| < d < r₁ + r₂.
|3 − 4| = 1 < 5 < 3 + 4 = 7. ✓ Circles intersect.

Step 4: Find the equation of the common chord.
Subtract Circle 1 from Circle 2:
(x − 4)² + (y − 3)² − 16 − (x² + y² − 9) = 0
x² − 8x + 16 + y² − 6y + 9 − 16 − x² − y² + 9 = 0
−8x − 6y + 18 = 0
4x + 3y − 9 = 0.

Step 5: Find the perpendicular distance from C₁ = (0, 0) to the chord 4x + 3y − 9 = 0.
p = |4(0) + 3(0) − 9|/√(16 + 9) = 9/5.

Step 6: Use the chord-length formula. If a chord is at distance p from the centre of a circle with radius r, the chord length is 2√(r² − p²).
Chord length = 2√(9 − (9/5)²) = 2√(9 − 81/25) = 2√((225 − 81)/25) = 2√(144/25) = 2 · 12/5 = 24/5.

Question 16

A circle with centre on the line 2x − y − 3 = 0 passes through the points (1, 2) and (3, 4). Find the radius of the circle.

Accepted Answer

Step 1: Let the centre be C = (h, k). Since C lies on 2x − y − 3 = 0:
2h − k − 3 = 0 → k = 2h − 3. ... (i)

Step 2: The circle passes through (1, 2) and (3, 4), so the distances from C to both points are equal (both equal to the radius).
(h − 1)² + (k − 2)² = (h − 3)² + (k − 4)². ... (ii)

Step 3: Expand equation (ii):
h² − 2h + 1 + k² − 4k + 4 = h² − 6h + 9 + k² − 8k + 16
−2h + 1 − 4k + 4 = −6h + 9 − 8k + 16
−2h − 4k + 5 = −6h − 8k + 25
4h + 4k = 20
h + k = 5. ... (iii)

Step 4: Solve (i) and (iii) simultaneously.
From (i): k = 2h − 3.
Substitute into (iii): h + 2h − 3 = 5 → 3h = 8 → h = 8/3.
Then: k = 2(8/3) − 3 = 16/3 − 9/3 = 7/3.
Centre: C = (8/3, 7/3).

Step 5: Find the radius using the distance from C to (1, 2):
r = √[(8/3 − 1)² + (7/3 − 2)²]
= √[(8/3 − 3/3)² + (7/3 − 6/3)²]
= √[(5/3)² + (1/3)²]
= √[25/9 + 1/9]
= √[26/9]
= √26/3.

Question 17

The circle x² + y² − 4x − 6y + k = 0 is orthogonal to the circle x² + y² − 2x − 4y − 7 = 0. Find the value of k.

Accepted Answer

Step 1: Recall the condition for two circles to be orthogonal.
Two circles are orthogonal if the tangents at their points of intersection are perpendicular. Equivalently, if circles are:
C₁: x² + y² + 2g₁x + 2f₁y + c₁ = 0
C₂: x² + y² + 2g₂x + 2f₂y + c₂ = 0
then they are orthogonal if and only if: 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Step 2: Identify coefficients for the given circles.
Circle 1: x² + y² − 4x − 6y + k = 0
Compare with x² + y² + 2g₁x + 2f₁y + c₁ = 0:
2g₁ = −4 → g₁ = −2
2f₁ = −6 → f₁ = −3
c₁ = k

Circle 2: x² + y² − 2x − 4y − 7 = 0
Compare with x² + y² + 2g₂x + 2f₂y + c₂ = 0:
2g₂ = −2 → g₂ = −1
2f₂ = −4 → f₂ = −2
c₂ = −7

Step 3: Apply the orthogonality condition.
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(−2)(−1) + 2(−3)(−2) = k + (−7)
2(2) + 2(6) = k − 7
4 + 12 = k − 7
16 = k − 7
k = 23.

Question 18

The circle x² + y² − 6x + 4y − 12 = 0 has center and radius respectively:

Accepted Answer

The general form of a circle is x² + y² + 2gx + 2fy + c = 0.

Given: x² + y² − 6x + 4y − 12 = 0

Step 1: Identify coefficients by comparing with general form.
2g = −6  ⟹  g = −3
2f = 4   ⟹  f = 2
c = −12

Step 2: Find the center using the formula: Center = (−g, −f).
Center = (−(−3), −2) = (3, −2)

Step 3: Find the radius using the formula: r = √(g² + f² − c).
r = √[(−3)² + (2)² − (−12)]
r = √[9 + 4 + 12]
r = √25
r = 5

Therefore, center = (3, −2) and radius = 5.

Question 19

The equation of the tangent to the circle x² + y² = 25 at the point (3, 4) is:

Accepted Answer

For a circle x² + y² = r², the equation of the tangent at point (x₀, y₀) on the circle is:
xx₀ + yy₀ = r²

Given: Circle x² + y² = 25, point (3, 4).

Step 1: Verify that (3, 4) lies on the circle.
3² + 4² = 9 + 16 = 25 ✓

Step 2: Apply the tangent formula.
Tangent equation: x(3) + y(4) = 25
3x + 4y = 25

Step 3: Verify by checking that the distance from center (0, 0) to the line 3x + 4y − 25 = 0 equals the radius.
Distance = |3(0) + 4(0) − 25| / √(3² + 4²) = 25 / √25 = 25 / 5 = 5 ✓

The tangent equation is 3x + 4y = 25.

Question 20

Two circles have equations x² + y² − 2x − 4y − 4 = 0 and x² + y² − 6x − 8y + 16 = 0. The distance between their centers is:

Accepted Answer

For a circle in general form x² + y² + 2gx + 2fy + c = 0, the center is (−g, −f).

Circle 1: x² + y² − 2x − 4y − 4 = 0
2g₁ = −2  ⟹  g₁ = −1
2f₁ = −4  ⟹  f₁ = −2
Center C₁ = (−g₁, −f₁) = (1, 2)

Circle 2: x² + y² − 6x − 8y + 16 = 0
2g₂ = −6  ⟹  g₂ = −3
2f₂ = −8  ⟹  f₂ = −4
Center C₂ = (−g₂, −f₂) = (3, 4)

Step 1: Find the distance between C₁(1, 2) and C₂(3, 4) using the distance formula.
d = √[(3 − 1)² + (4 − 2)²]
d = √[2² + 2²]
d = √[4 + 4]
d = √8
d = 2√2

The distance between the centers is 2√2.

Agniveer (All) Circles

Circles— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Circles — Practice Questions

60-Second Revision — Circles