Two circles of radii 10 cm and 5 cm touch each other externally at a point A. PQ Is the direct common tangent of those two circles of centres O1 and O2 , respectively. The length of PQ is equal to:
10√2 cm
Calculating the Length of a Direct Common Tangent
This problem involves two circles that touch each other externally. We are given their radii and asked to find the length of a direct common tangent between them. A direct common tangent is a line segment that is tangent to both circles on the same side.
Understanding Externally Touching Circles
When two circles touch each other externally, the distance between their centres is equal to the sum of their radii.
- Radius of the first circle, $r_1 = 10$ cm (centre $O_1$).
- Radius of the second circle, $r_2 = 5$ cm (centre $O_2$).
- The circles touch externally.
The distance between the centres $O_1$ and $O_2$ is:
Distance ($d$) $= r_1 + r_2 = 10$ cm $+ 5$ cm $= 15$ cm.
Formula for the Length of a Direct Common Tangent
The length of a direct common tangent (L) between two circles with radii $r_1$ and $r_2$ and the distance between their centres $d$ is given by the formula:
$\text{L} = \sqrt{d^2 - (r_1 - r_2)^2}$
Step-by-Step Calculation of Tangent Length
We have $r_1 = 10$ cm, $r_2 = 5$ cm, and $d = 15$ cm. Let's plug these values into the formula:
- Calculate the difference in radii: $r_1 - r_2 = 10 - 5 = 5$ cm.
- Square the distance between centres: $d^2 = 15^2 = 225$.
- Square the difference in radii: $(r_1 - r_2)^2 = 5^2 = 25$.
- Subtract the squared difference in radii from the squared distance: $d^2 - (r_1 - r_2)^2 = 225 - 25 = 200$.
- Take the square root of the result to find the length of the tangent: $\text{L} = \sqrt{200}$.
To simplify $\sqrt{200}$, we look for perfect square factors of 200:
$\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2}$ cm.
So, the length of the direct common tangent PQ is $10\sqrt{2}$ cm.
Summary of Results
| Parameter | Value |
|---|---|
| Radius 1 ($r_1$) | 10 cm |
| Radius 2 ($r_2$) | 5 cm |
| Distance between centres ($d$) | 15 cm (since touching externally) |
| Difference in radii ($r_1 - r_2$) | 5 cm |
| Length of Direct Common Tangent (L) | $\sqrt{d^2 - (r_1 - r_2)^2} = 10\sqrt{2}$ cm |
Revision Table: Circles and Tangents
| Concept | Description | Formula (if applicable) |
|---|---|---|
| Circles touching externally | Distance between centres equals sum of radii. | $d = r_1 + r_2$ |
| Circles touching internally | Distance between centres equals difference of radii. | $d = |r_1 - r_2|$ |
| Direct Common Tangent | Tangent line on the same side of both circles. | $\text{L} = \sqrt{d^2 - (r_1 - r_2)^2}$ |
| Transverse Common Tangent | Tangent line that crosses between the circles. | $\text{L} = \sqrt{d^2 - (r_1 + r_2)^2}$ |
Additional Information: Deriving the Tangent Length Formula
The formula for the length of a direct common tangent can be derived using geometry. Imagine drawing a line through the centre of the smaller circle ($O_2$) parallel to the tangent PQ, meeting the radius $O_1P$ (extended if necessary) at a point, say R. This creates a rectangle $O_2QPR$ (if $P$ and $Q$ are the points of tangency) and a right-angled triangle $O_1RO_2$. The length $RQ$ is equal to $O_2Q = r_2$. The length $RP$ is parallel to $O_2Q$ and equal to $r_2$. Then $O_1R$ would be the difference in radii, $|r_1 - r_2|$. The distance $O_1O_2$ is the hypotenuse $d$. The length of the tangent PQ is equal to $RO_2$. By the Pythagorean theorem in triangle $O_1RO_2$, we have $O_1O_2^2 = O_1R^2 + RO_2^2$. Substituting the values, $d^2 = (r_1 - r_2)^2 + \text{PQ}^2$. Rearranging gives $\text{PQ}^2 = d^2 - (r_1 - r_2)^2$, and thus $\text{PQ} = \sqrt{d^2 - (r_1 - r_2)^2}$. This confirms the formula used.
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