CAT — Circles

Given:
A chord of length 10√3 m subtends an angle of 120° at the center of a circle.
Find the area of the smaller region cut off by the chord.

Step 1: Find the radius of the circle
For a chord,
Chord length = 2r sin(θ/2)
Here,
10√3 = 2r sin 60°
= 2r × √3/2
= r√3
Therefore,
r = 10 m

Step 2: Find the area of the sector
The smaller region corresponds to the sector of angle 120°.
Area of the sector
= (120/360) × π × 10²
= 100π/3

Step 3: Find the area of the triangle
The triangle formed by the two radii and the chord has
two sides = 10 m
included angle = 120°.
Area
= (1/2) × 10 × 10 × sin 120°
= 50 × √3/2
= 25√3

Step 4: Find the area of the smaller region
Area of the smaller region
= Area of sector − Area of triangle
= 100π/3 − 25√3
= 25(4π/3 − √3)

Final Answer
Answer:
D. 25(4π/3 − √3)

Given:
Three equal circles of radius r touch each other externally.
Two more circles, X and Y, are drawn such that each touches all the three circles.
Radius of X is greater than that of Y.
Find the ratio of the radii of X and Y.

Step 1: Locate the centres
The centres of the three equal circles form an equilateral triangle of side
2r.
The centre of both circles X and Y is the common centre (circumcentre) of this equilateral triangle.
The distance from this centre to each vertex is the circumradius of the equilateral triangle.
Circumradius
= (2r)/√3
= 2r/√3

Step 2: Find the radius of the larger circle X
Since X encloses the three circles, the distance between its centre and the centre of each small circle is
R − r,
where R is the radius of X.
Hence,
R − r = 2r/√3
R = r + 2r/√3
= r(1 + 2/√3)

Step 3: Find the radius of the smaller circle Y
Since Y lies inside the gap, the distance between its centre and the centre of each small circle is
r + y,
where y is the radius of Y.
Hence,
r + y = 2r/√3
y = 2r/√3 − r
= r(2/√3 − 1)

Step 4: Find the ratio
R : y
= (1 + 2/√3) : (2/√3 − 1)
Multiply both terms by √3,
= (√3 + 2) : (2 − √3)
Rationalizing,
= (√3 + 2)² : (4 − 3)
= (3 + 4 + 4√3) : 1
= (7 + 4√3) : 1

Answer:
A. (7 + 4√3) : 1

Step 1: Rewrite the second inequality
Complete the squares:
x² + y² + 4x − 4y = (x+2)² − 4 + (y−2)² − 4 = (x+2)² + (y−2)² − 8
Hence: −4 ≤ (x+2)² + (y−2)² − 8 ≤ 0
Adding 8: 4 ≤ (x+2)² + (y−2)² ≤ 8
This is the annular region between two concentric circles with centre (−2, 2), inner radius 2, outer radius 2√2.

Step 2: Interpret the line
The line y = x + 4. Substituting the centre (−2, 2): 2 = −2 + 4 ✓
The line passes through the centre, dividing the annular region into two equal halves.
The required region satisfies y ≥ x + 4, so it is half the annulus.

Step 3: Find the required area
Area of annulus = π[(2√2)² − 2²] = π(8 − 4) = 4π
Required area = (1/2) × 4π = 2π

Step 1: Identify the quadrilateral formed by the tangents. Since PQ, PR and AB are tangents to the circle, the radius is perpendicular to the tangent at the point of contact. Let the points of contact of the tangents PQ, PR and AB be X, Y and T respectively. Then, OX ⟂ PQ, OY ⟂ PR, OT ⟂ AB.

Step 2: Use the property of tangents from an external point. Since A is an external point, the two tangents from A are AX and AT. Therefore, OA bisects ∠XOT. Similarly, OB bisects ∠YOT. Given ∠AOB = 50°. Hence, ∠XOY = 2 × ∠AOB = 2 × 50° = 100°.

Step 3: Use the angle between two tangents. The angle between two tangents drawn from an external point is supplementary to the central angle subtended by the points of contact. Therefore, ∠APB = 180° − ∠XOY = 180° − 100° = 80°.

Given:
A regular octagon has side length
6 cm.
Points A, C, E and G form a square.
Find the area of square ACEG.

Step 1: Find the circumradius of the octagon
For a regular octagon,
Side = 2R sin 22.5°
So,
6 = 2R sin 22.5°
Using,
sin 22.5° = √(2 − √2)/2,
6 = R√(2 − √2)
Therefore,
R = 6/√(2 − √2)

Step 2: Find the side of square ACEG
Vertices A and C are separated by a central angle of
90°.
Hence,
AC is the chord subtending 90°.
Therefore,
AC = R√2
Substituting the value of R,
AC = 6√2/√(2 − √2)
Square both sides,
AC² = 72/(2 − √2)
Rationalizing,
AC² = 72(2 + √2)/2
= 36(2 + √2)
Since AC is the side of the square,
Area of the square
= AC²
= 36(2 + √2)

Final Answer
Answer:
D. 36(2 + √2)

Since one side is the diameter, the angle opposite it is 90° (Thales' theorem).
The triangle is right-angled with perpendicular sides λa and λb.
Hypotenuse = 2r (diameter).

Using Pythagoras:
(λa)² + (λb)² = (2r)²
λ²(a² + b²) = 4r² → λ² = 4r²/(a² + b²)

Area = (1/2) × λa × λb = (1/2) × λ²ab
= (1/2) × (4r²/(a² + b²)) × ab
= 2abr²/(a² + b²)

Given, x² + y² + 4x − 6y − 3 = 0
Complete the squares, (x + 2)² + (y − 3)² = 16
Hence,
● Centre of the circle C = (−2, 3)
● Radius = 4
Let P be the point of intersection of the two tangents.
Suppose the distance of P from the centre is d.
In the right triangle formed by the centre, the point of contact, and P,
sin(angle between OP and a tangent) = Radius / OP = 4/d
The angle between the two tangents is 2 × angle between OP and one tangent
Given, 2 × angle = 60°
So, angle = 30°
Therefore, sin 30° = 4/d → 1/2 = 4/d → d = 8
Hence, the locus L is the circle
Centre = (−2, 3)
Radius = 8
Its equation is (x + 2)² + (y − 3)² = 64
The line x = 6 is vertical.
The circle touches this line at its rightmost point.
Rightmost point = (Centre x-coordinate + Radius, Centre y-coordinate) = (−2 + 8, 3) = (6, 3)
Answer:
B. (6, 3)

Radius r = 6√2. Let M be midpoint of PQ. △QCM is right-angled with ∠MQC = 45°, so it is 45°–45°–90°. CM = CQ/√2 = 6√2/√2 = 6 cm. So distance of PQ from C = 6 cm.

Distance of SR from C = (2/3)×6 = 4 cm.

Length of PQ = 2√(72−36) = 2√36 = 12 cm. Length of SR = 2√(72−16) = 2√56 = 4√14 cm.

Since chords are on opposite sides of the diameter, distance between them = 6+4 = 10 cm.

Area of trapezium PQRS = 1/2 × (12 + 4√14) × 10 = 5(12 + 4√14) = 20(3 + √14).

For a cyclic quadrilateral, if the diagonals intersect at E, then
AE/CE = (AB × AD)/(CB × CD)
(This is a standard result obtained using similar triangles.)
Given, AB : CD = 2 : 1, BC : AD = 5 : 4
Let, AB = 2k, CD = k, BC = 5m, AD = 4m
Now, AE/CE = (AB × AD)/(BC × CD) = (2k × 4m)/(5m × k) = 8/5
Therefore, AE : CE = 8 : 5
Answer:
C. 8 : 5

Step 1: △ABC is isosceles with AB = AC = 12 cm. Let BC = b.
Midpoint of BC is M; AM² = 144 − b²/4.

Step 2: AD = 8 cm.
Using the condition ∠AEB = ∠ACB, points B, C, D, E are concyclic
(since ∠AEB = ∠ACB means E lies on the circumcircle of △ABC, or by
power of a point / similar triangles).
By Power of a Point: AD × AE = AB × AC
8 × AE = 12 × 12
8 × AE = 144
AE = 18 cm