Circles, Semi Circles and Tangents — NCERT Solutions

Given: Circle of radius $R = 20$ mm; point P on the circle.

Concept: The tangent at any point on a circle is perpendicular to the radius drawn to that point.

Steps of Construction:

1. Draw a circle with centre O and radius $R = 20$ mm.
2. Mark any point P on the circumference.
3. Join OP and extend it beyond P (this is the normal at P).
4. At point P, draw a line perpendicular to OP.
5. This perpendicular line is the required tangent at P.

Result: The line drawn perpendicular to OP at point P is the tangent to the circle at P. The angle between the tangent and radius OP is $90°$.

Given: Circle with centre O, radius $R = 25$ mm; external point P at distance $OP = 55$ mm from O.

Concept: From an external point, two tangents can be drawn to a circle. The method uses the fact that the angle in a semicircle is $90°$ (tangent is perpendicular to radius at point of contact).

Steps of Construction:

1. Draw circle with centre O and radius $R = 25$ mm.
2. Mark external point P such that $OP = 55$ mm.
3. Join OP. Find the midpoint M of OP.
4. With M as centre and MO (= MP = $\frac{55}{2} = 27.5$ mm) as radius, draw a semicircle on OP.
5. This semicircle intersects the given circle at points T$_1$ and T$_2$.
6. Join P–T$_1$ and P–T$_2$. These are the two required tangents.

Verification: $\angle$OT$_1$P = $\angle$OT$_2$P = $90°$ (angle in semicircle), confirming OT$_1$ $\perp$ PT$_1$ and OT$_2$ $\perp$ PT$_2$.

Result: PT$_1$ and PT$_2$ are the two tangents from external point P to the circle.

Length of tangent:
$$PT = \sqrt{OP^2 - R^2} = \sqrt{55^2 - 25^2} = \sqrt{3025 - 625} = \sqrt{2400} \approx 48.99 \text{ mm}$$

Given: Two circles, each with radius $R = 25$ mm; distance between centres $OO_1 = 65$ mm.

Concept: For two circles of equal radii, the external common tangents are parallel to the line joining the centres and are drawn by erecting perpendiculars at each centre.

Steps of Construction:

1. Draw two circles each of radius $R = 25$ mm with centres O and O$_1$ such that $OO_1 = 65$ mm.
2. At centre O, erect a perpendicular to $OO_1$. Mark point P on the circle where this perpendicular meets the circle (above $OO_1$).
3. At centre O$_1$, erect a perpendicular to $OO_1$ on the same side. Mark point Q on the circle.
4. Join P and Q. PQ is the first external common tangent.
5. Repeat on the other side (below $OO_1$) to get points P$'$ and Q$'$.
6. Join P$'$Q$'$. This is the second external common tangent.

Result: PQ and P$'$Q$'$ are the two external (direct) common tangents to the two equal circles.

Given: Two circles, each radius $R = 30$ mm, touching externally.

Concept: When two circles touch externally, distance between centres $= R_1 + R_2 = 30 + 30 = 60$ mm. For equal radii, external tangents are perpendicular to $OO_1$.

Steps of Construction:

1. Mark centres O and O$_1$ such that $OO_1 = 60$ mm.
2. Draw both circles of radius 30 mm. They will touch externally at the midpoint of $OO_1$.
3. At O, draw a perpendicular to $OO_1$; it meets the circle at P (above) and P$'$ (below).
4. At O$_1$, draw a perpendicular to $OO_1$; it meets the circle at Q (above) and Q$'$ (below).
5. Join PQ → first external common tangent.
6. Join P$'$Q$'$ → second external common tangent.

Note: Since the circles touch externally at one point, an internal common tangent also exists at the point of contact (common internal tangent is the common tangent at the point of touch).

Result: PQ and P$'$Q$'$ are the two required external common tangents.

Given: Two circles with $R_1 = 20$ mm and $R_2 = 15$ mm touching externally.

Concept: Distance between centres $= R_1 + R_2 = 20 + 15 = 35$ mm. For unequal radii, the external tangent is found using the method of Example 7 (Fig. 2.9).

Steps of Construction:

1. Mark centres O and O$_1$ with $OO_1 = 35$ mm.
2. Draw circle of radius 20 mm at O and circle of radius 15 mm at O$_1$. They touch externally at point A (midpoint dividing $OO_1$ in ratio 20:15).
3. Draw a semicircle on $OO_1$ as diameter.
4. At point A (point of external contact), erect a perpendicular to $OO_1$. Let it meet the semicircle at point B.
5. With B as centre and BA as radius, draw an arc cutting the two given circles at points P (on larger circle) and Q (on smaller circle).
6. Join PQ. This is the required external common tangent.

Result: PQ is the external common tangent to the two touching circles of radii 20 mm and 15 mm.

Given: Circle 1: centre O, radius $R_1 = 30$ mm; Circle 2: centre O$_1$, radius $R_2 = 15$ mm; $OO_1 = 70$ mm.

Concept: For two circles of unequal radii, the external common tangent is found by the following method. The external centre of similitude divides $OO_1$ externally in the ratio $R_1 : R_2$.

Steps of Construction:

1. Draw the two circles with centres O and O$_1$, $OO_1 = 70$ mm, radii 30 mm and 15 mm respectively.
2. Draw a third auxiliary circle with centre O and radius equal to the difference of radii: $R_1 - R_2 = 30 - 15 = 15$ mm.
3. From O$_1$, draw a tangent to this auxiliary circle of radius 15 mm (using the semicircle method: find midpoint M of $OO_1$, draw semicircle of radius $MO_1$, which intersects auxiliary circle at T).
4. The line O$_1$T, when extended, gives the direction of the tangent. Draw a line through O parallel to O$_1$T meeting the original circle at P.
5. Through O$_1$, draw a line parallel to OP meeting the smaller circle at Q.
6. Join PQ. PQ is the required external common tangent.

Alternative (direct method):

1. Locate external centre of similitude S on line $OO_1$ extended beyond O$_1$, dividing $OO_1$ externally in ratio $30:15 = 2:1$.

$$SO = \frac{R_1}{R_1 - R_2} \times OO_1 = \frac{30}{15} \times 70 = 140 \text{ mm from O}$$
So S is 70 mm beyond O$_1$.

2. From S, draw tangents to either circle (using semicircle method). The tangent lines are the external common tangents.

Result: The external common tangent PQ is drawn to the two circles.

Given: Two intersecting circles; centre distance $OO_1 = 30$ mm; $R_1 = 25$ mm, $R_2 = 15$ mm.

Check for intersection: $R_1 + R_2 = 40$ mm $> 30$ mm and $|R_1 - R_2| = 10$ mm $< 30$ mm. ✓ Circles intersect.

Concept: Even for intersecting circles, external common tangents exist (two of them). The method uses the external centre of similitude.

Steps of Construction:

1. Draw circle with centre O, radius 25 mm and circle with centre O$_1$, radius 15 mm, with $OO_1 = 30$ mm. The circles will intersect at two points.
2. Draw auxiliary circle with centre O and radius $= R_1 - R_2 = 25 - 15 = 10$ mm.
3. From O$_1$, draw a tangent to this auxiliary circle:

• Find midpoint M of $OO_1$.
• With M as centre and $MO_1$ as radius, draw a semicircle intersecting the auxiliary circle at point T.
• O$_1$T is the tangent direction.

4. Through O, draw a line parallel to O$_1$T. It meets the larger circle at point P.
5. Through O$_1$, draw a line parallel to OP meeting the smaller circle at Q (on the same side).
6. PQ is the required external common tangent.
7. Repeat on the other side to get the second external common tangent P$'$Q$'$.

Result: Two external common tangents are drawn to the two intersecting circles.

Given: Two circles, each radius $R = 30$ mm; $OO_1 = 80$ mm.

Concept: The internal common tangent passes between the two circles and crosses the line $OO_1$ at the internal centre of similitude (midpoint for equal circles). The tangent is perpendicular to the line joining the midpoint to the point of tangency.

Steps of Construction:

1. Draw two circles each of radius 30 mm with centres O and O$_1$, $OO_1 = 80$ mm.
2. Find midpoint M of $OO_1$ (since radii are equal, M is the internal centre of similitude).
3. With M as centre and MO (= 40 mm) as radius, draw a semicircle.
4. With M as centre and radius $R = 30$ mm, draw an arc cutting the semicircle at point T.
5. Draw a line through M and T. This line intersects the two circles at points P and Q.
6. PQ is the required internal common tangent.

Alternative method:

1. With M as centre and MO = 40 mm as radius, draw a circle.
2. From M, draw a perpendicular to $OO_1$. The perpendicular meets the circle at a point. From this point, draw the tangent to either given circle — this gives the internal tangent.

Simpler direct method for equal circles:

1. M = midpoint of $OO_1$.
2. At M, erect a perpendicular to $OO_1$.
3. From M, mark off 30 mm (= R) on both sides of $OO_1$ along this perpendicular to get P and Q.
4. PQ is the internal common tangent (it passes through M and is perpendicular to $OO_1$).

Result: The line through M perpendicular to $OO_1$ is the internal common tangent. A second internal tangent can be drawn similarly on the other side.

Given: Circle 1: centre O, $R_1 = 25$ mm; Circle 2: centre O$_1$, $R_2 = 20$ mm; $OO_1 = 70$ mm.

Concept: The internal centre of similitude divides $OO_1$ internally in the ratio $R_1 : R_2 = 25 : 20 = 5 : 4$.

Steps of Construction:

1. Draw the two circles with given dimensions.
2. Locate internal centre of similitude M on $OO_1$:

$$OM = \frac{R_1}{R_1 + R_2} \times OO_1 = \frac{25}{45} \times 70 = \frac{1750}{45} \approx 38.9 \text{ mm from O}$$

3. With M as centre and MO as radius, draw a semicircle on $OO_1$.
4. With M as centre and $R_1 = 25$ mm as radius, draw an arc cutting the semicircle at point T.
5. Join MT and extend. This line meets the two circles at points P and Q.
6. PQ is the required internal common tangent.

Verification: $\angle$OPM = $90°$ (angle in semicircle), so OP $\perp$ PQ, confirming PQ is tangent to circle 1. Similarly for circle 2.

Result: PQ is the internal common tangent to the two given circles.

Given: Equilateral triangle with height $h = 55$ mm.

Concept: The inscribed circle (incircle) of a triangle has its centre at the incentre (intersection of angle bisectors). For an equilateral triangle, the incentre coincides with the centroid and circumcentre.

Finding side from height:
$$h = \frac{\sqrt{3}}{2} \times a \implies a = \frac{2h}{\sqrt{3}} = \frac{2 \times 55}{1.732} \approx 63.5 \text{ mm}$$

Steps of Construction:

1. Draw a vertical line of length 55 mm. Mark the apex A at top and foot F at bottom.
2. At F, draw a horizontal line. With F as centre and $a/2 \approx 31.75$ mm on each side, mark B and C.
3. Join AB and AC to complete the equilateral triangle ABC.
4. Draw the angle bisector from vertex A (which is also the median and altitude for equilateral triangle).
5. Draw the angle bisector from vertex B.
6. Let these bisectors meet at O. This is the incentre.
7. From O, drop a perpendicular to side BC. Let the foot be G. The length OG is the inradius $r$.

$$r = \frac{h}{3} = \frac{55}{3} \approx 18.3 \text{ mm}$$

8. With O as centre and OG as radius, draw the inscribed circle.

Result: The circle with centre O and radius $\approx 18.3$ mm is inscribed in the equilateral triangle.

Given: Square ABCD with side $= 40$ mm.

Concept: The inscribed circle of a square has its centre at the intersection of the diagonals and its radius equal to half the side length.

Steps of Construction:

1. Draw square ABCD with each side $= 40$ mm.
2. Draw both diagonals AC and BD. They intersect at centre O.
3. From O, draw a perpendicular to any side (say AB). Let the foot be M. Then $OM = 20$ mm (half the side).
4. With O as centre and $OM = 20$ mm as radius, draw the inscribed circle.

Inradius:
$$r = \frac{\text{side}}{2} = \frac{40}{2} = 20 \text{ mm}$$

Result: The circle with centre O and radius 20 mm is inscribed in the square, touching all four sides.

Given: Rhombus with diagonals $d_1 = 70$ mm and $d_2 = 40$ mm.

Concept: The diagonals of a rhombus bisect each other at right angles. The incentre is at the intersection of the diagonals. The inradius equals the perpendicular distance from the centre to any side.

Finding side of rhombus:
$$a = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} = \sqrt{35^2 + 20^2} = \sqrt{1225 + 400} = \sqrt{1625} \approx 40.3 \text{ mm}$$

Steps of Construction:

1. Draw diagonal AC $= 70$ mm horizontally.
2. Find midpoint O of AC.
3. At O, draw a perpendicular. Mark B and D at 20 mm each side of O on this perpendicular.
4. Join AB, BC, CD, DA to complete the rhombus.
5. The diagonals intersect at O, which is the incentre.
6. From O, drop a perpendicular to side AB. Let foot be G. Measure OG — this is the inradius $r$.

Inradius calculation:
$$\text{Area of rhombus} = \frac{d_1 \times d_2}{2} = \frac{70 \times 40}{2} = 1400 \text{ mm}^2$$
$$\text{Perimeter} = 4a = 4 \times 40.3 = 161.2 \text{ mm}$$
$$r = \frac{\text{Area}}{\text{Semi-perimeter}} = \frac{1400}{80.6} \approx 17.4 \text{ mm}$$

7. With O as centre and OG ($\approx 17.4$ mm) as radius, draw the inscribed circle.

Result: The circle with centre O and radius $\approx 17.4$ mm is inscribed in the rhombus, touching all four sides.

Given: Regular pentagon with side $= 45$ mm.

Concept: The inscribed circle of a regular polygon has its centre at the centre of the polygon (intersection of diagonals/angle bisectors) and radius equal to the apothem (perpendicular distance from centre to a side).

Steps of Construction:

1. Draw a regular pentagon ABCDE with each side $= 45$ mm (using the standard method: draw base AB = 45 mm, construct interior angles of $108°$ at each vertex).
2. Draw any two diagonals (e.g., AC and BD). Their intersection gives centre O.
3. From O, drop a perpendicular to side AB. Let the foot be M. OM is the apothem (inradius).
4. With O as centre and OM as radius, draw the inscribed circle.

Apothem calculation:
$$r = \frac{a}{2} \cot\left(\frac{\pi}{5}\right) = \frac{45}{2} \times \cot 36° = 22.5 \times 1.376 \approx 30.9 \text{ mm}$$

Result: The circle with centre O and radius $\approx 30.9$ mm is inscribed in the regular pentagon, touching all five sides.

Given: Regular hexagon with diagonal (longest) $= 70$ mm, so side $= 35$ mm (since diagonal of regular hexagon $= 2 \times$ side).

Concept: Same as Example 16. The inscribed circle centre is at the intersection of opposite vertex diagonals, and radius equals the perpendicular from centre to a side.

Steps of Construction:

1. Draw regular hexagon ABCDEF with diagonal $AD = 70$ mm (side $= 35$ mm).
2. Join opposite corners: draw diagonals AD, BE and CF. They all meet at centre O.
3. From O, drop a perpendicular OG to side AB.
4. With O as centre and OG as radius, draw the inscribed circle.

Inradius:
$$r = \frac{\sqrt{3}}{2} \times \text{side} = \frac{\sqrt{3}}{2} \times 35 = \frac{1.732 \times 35}{2} \approx 30.3 \text{ mm}$$

Result: The circle with centre O and radius $\approx 30.3$ mm is inscribed in the regular hexagon.

Given: Regular octagon with side $= 25$ mm.

Concept: Same as Example 17. The inscribed circle (incircle) has its centre at the intersection of opposite-vertex diagonals and radius equal to the apothem.

Steps of Construction:

1. Draw a regular octagon ABCDEFGH with each side $= 25$ mm (interior angle $= 135°$).
2. Join any two opposite corners (e.g., AE and BF). Their intersection gives centre O.
3. From O, drop a perpendicular OK to side AB.
4. With O as centre and OK as radius, draw the inscribed circle.

Apothem (inradius):
$$r = \frac{a}{2}(1 + \sqrt{2}) = \frac{25}{2}(1 + 1.414) = 12.5 \times 2.414 \approx 30.2 \text{ mm}$$

Result: The circle with centre O and radius $\approx 30.2$ mm is inscribed in the regular octagon, touching all eight sides.

Given: Angle $\angle ABC = 60°$; $AB = BC = 80$ mm; circle radius $r = 15$ mm touching both arms AB and BC.

Concept: A circle touching both arms of an angle has its centre on the angle bisector of that angle, at a perpendicular distance equal to the radius from each arm.

Steps of Construction:

1. Draw ray BA of length 80 mm.
2. At B, construct an angle of $60°$ and draw ray BC of length 80 mm.
3. Draw the angle bisector of $\angle ABC$ (bisect the $60°$ angle to get a $30°$ ray from B).
4. The centre O of the required circle lies on this bisector.
5. The perpendicular distance from O to line AB (or BC) must equal $r = 15$ mm.
6. Draw a line parallel to AB at a perpendicular distance of 15 mm (on the interior side of the angle).
7. The intersection of this parallel line with the angle bisector gives centre O.
8. With O as centre and radius $= 15$ mm, draw the required circle.

Distance of O from B:
$$BO = \frac{r}{\sin(\angle ABC / 2)} = \frac{15}{\sin 30°} = \frac{15}{0.5} = 30 \text{ mm}$$

So O is 30 mm from B along the angle bisector.

Result: The circle with centre O (30 mm from B on the angle bisector) and radius 15 mm touches both lines AB and BC.

Given: Pulley 1: centre O, radius $R_1 = 30$ mm; Pulley 2: centre O$_1$, radius $R_2 = 20$ mm; $OO_1 = 70$ mm.

(i) Direct Belt — Two External Common Tangents:

Concept: In a direct (open) belt drive, the belt does not cross between pulleys. The belt runs as external common tangents to both pulleys.

Steps of Construction:

1. Draw circle with centre O, radius 30 mm and circle with centre O$_1$, radius 20 mm, with $OO_1 = 70$ mm.
2. Draw auxiliary circle with centre O and radius $= R_1 - R_2 = 30 - 20 = 10$ mm.
3. From O$_1$, draw a tangent to this auxiliary circle:

• Midpoint M of $OO_1$; with M as centre and $MO_1 = 35$ mm as radius, draw semicircle intersecting auxiliary circle at T.
• O$_1$T is the tangent to auxiliary circle.

4. Through O, draw line parallel to O$_1$T meeting the larger circle at P (above $OO_1$).
5. Through O$_1$, draw line parallel to OP meeting smaller circle at Q (above $OO_1$).
6. PQ is the first external (direct) common tangent (upper belt).
7. Repeat below $OO_1$ to get P$'$Q$'$ as the second external common tangent (lower belt).

Result for (i): PQ and P$'$Q$'$ represent the direct belt arrangement.

(ii) Cross Belt — Two Internal Common Tangents:

Concept: In a cross belt drive, the belt crosses between the pulleys. The belt runs as internal common tangents to both pulleys.

Steps of Construction:

1. Using the same two circles.
2. Locate internal centre of similitude M$'$ on $OO_1$, dividing it internally in ratio $R_1 : R_2 = 30 : 20 = 3 : 2$:

$$OM' = \frac{R_1}{R_1 + R_2} \times OO_1 = \frac{30}{50} \times 70 = 42 \text{ mm from O}$$

3. With M$'$ as centre and $M'O = 42$ mm as radius, draw a semicircle.
4. With M$'$ as centre and $R_1 = 30$ mm as radius, draw an arc cutting the semicircle at point T$_1$.
5. Join M$'$T$_1$ and extend to meet the two circles at points P$_1$ and Q$_1$.
6. P$_1$Q$_1$ is the first internal common tangent (upper cross belt).
7. Repeat on the other side to get P$_2$Q$_2$ as the second internal common tangent (lower cross belt).

Result for (ii): P$_1$Q$_1$ and P$_2$Q$_2$ represent the cross-belt arrangement.

Summary Table:

Result: Both belt arrangements are drawn showing the two pulleys connected by direct and cross belts respectively.