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Common Tangents
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CAT 2025 Lesson : Circles - Common Tangents

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9. Common Tangents

Properties Figure
When one circle is inside the other, then there can be no common tangent.
When two circles touch internally at exactly one point, then there can be a maximum of 1 common tangent.

The two circles touch at A. The common tangent that can be drawn is aaa
When two circles intersect each other at 222 points, then 2\bm{2}2 common tangents can be drawn.
When two circles touch externally at exactly one point, then there can be a maximum of 3\bm{3}3 common tangents.

Here, the two circles touch at A. The common tangents that can be drawn are a,ba, ba,b and ccc
When two non- overlapping circles do not touch each other, then there can be a maximum of 4\bm{4}4 common tangents.
The tangents that do not intersect each other are called direct tangents and the tangents that do not intersect each other are called transverse tangents.

In the two circles,
AB and CD are the direct common tangents
EF and GH are the transverse common tangents

Where r1r_1r1​ & r2r_2r2​ are the radii of the circles with O & P as centres,
AB2^{2}2 === CD2^{2}2 === OP2−(r1−r2)2^{2} - (r_1 - r_2)^{2}2−(r1​−r2​)2
EF2^{2}2 === GH2^{2}2 === OP2−(r1+r2)2^{2} - (r_1 + r_2)^{2}2−(r1​+r2​)2

10. Semicircle

Area of a semicircle === πr22\dfrac{\pi r^{2}}{2}2πr2​ and Circumference of a semicircle === πr\pi rπr

Example 15

What is the minimum possible ratio of the area of a semicircle to the area of the largest square that can be inscribed in the semicircle, such that one of the sides of the square lies on the diameter of the semicircle?

(1) 5π:45 \pi : 45π:4            (2) 5π:85 \pi : 85π:8            (3) 4π:54 \pi : 54π:5            (4) 8π:58 \pi : 58π:5           

Solution

To minimise the ratio, the area of the square which is in the denominator should be maximised.

If one of the sides is on the diameter of the circumcircle, the area of the square will be maximum when the other two vertices lie on the arc of the semicircle.

Let CD lie on the diameter, while A and B lie on the arc. Where rrr is the radius and sss is the side of the square, OA === rrr and AD === sss

△\triangle△OAD ≅ △\triangle△OBC (RHS congruency)
∴\therefore∴ OC === OD === s2\dfrac{s}{2}2s​

Applying pythagoras theorem, r2=s2+(s2)2r^{2} = s^{2} + \left( \dfrac{s}{2} \right)^{2}r2=s2+(2s​)2 = 5s24\dfrac{5 s^{2}}{4}45s2​

Ratio of area of semicircle to square === πr22:s2\dfrac{\pi r^{2}}{2} : s^{2}2πr2​:s2 === π5s28:s2\pi \dfrac{5 s^{2}}{8} : s^{2}π85s2​:s2 === 5π:85 \pi : 85π:8

Answer: (2) 5π:85 \pi : 85π:8


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