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*Result \(4\)*:

If two circles touch externally, the distance between their centres equals the sum of their radii.

**Explanation**:

If two circles touch externally at a point \(B\), then the distance \(OA\) is equal to the sum of the radii \(OB\) and \(AB\).

\(\Rightarrow\) \(OA\) \(=\) \(OB\) \(+\) \(AB\)

*Proof of the result*:

Let the two circles with centres \(O\) and \(A\) intersect each other externally at point \(B\).

Let the radius \(OB\) \(=\) \(r_{1}\) and \(AB\) \(=\) \(r_{2}\) and \(r_{1}\) \(>\) \(r_{2}\).

Let the distance between the centres be \(d\).

\(\Rightarrow\) \(OA\) \(=\) \(d\)

From the figure, we observe that \(OA\) \(=\) \(OB\) \(+\) \(AB\).

\(\Rightarrow\) \(d\) \(=\) \(r_{1}\) \(+\) \(r_{2}\)

Example:

Two circle with radii \(4\) \(cm\) and \(5\) \(cm\) intersect at a point \(O\) externally. If so, find the distance between their centres.

**Solution**:

By the result, we know that:

Distance between the centres \(=\) Sum of the radii.

Thus, the distance between the centres \(=\) \(4\) \(cm\) \(+\) \(5\) \(cm\)

\(=\) \(9\) \(cm\)