Chapter 10.4, Problem 21WE

To determine

Construct a common internal tangent to circles with centre P and Q.

Explanation of Solution

Given:

There are two circles with centre P and Q respectively. They do not intersect and Q is not inside circle with centre P.The radii of circle with centre P and Q are p and q respectively, with p > q .

Calculation:

Draw the two circles with centres P and Q and radii p and q respectively , with p > q .

Draw a tangent $\overleftrightarrow{XY}$ :

1. Form an auxiliary circle P’ with centre P and radius  p + q .
2. Construct a tangent to circle P’ from Q , and call the point of tangency Z.
3. Draw $\overrightarrow{PZ}$ . $\overrightarrow{PZ}$ intersects circle with centre P at point X.
4. With center X and radius ZQ , draw an arc that intersects circle with centre Q in a point Y.
5. Draw $\overleftrightarrow{XY}$ .

  

Justification that $\overleftrightarrow{XY}$ is a tangent to the circles:

Draw $\overline{QY}$ and $\overline{PZ}\parallel \overline{QY}$ .

Tangents to a circle are perpendicular to the radii at the points of contact.Thus , in the above diagram , since $\overline{QZ}$ is tangent to circle P’ , so , $\overline{PZ}\perp \overline{QZ}$ .

By construction , $ZX=QY=q$ and $\overline{ZX}\parallel \overline{QY}$ and $m\angle XZQ=90°$ .

Hence XZQY is a rectangle.

So, $m\angle ZXY=90°\Rightarrow m\angle PXY=90°\mathrm{.............}[\text{linear angles]}$

Hence the radii $\overline{PX}\text{ and }\overline{QY}$ are perpendicular to $\overleftrightarrow{XY}$ and $\overline{PQ}$ doe intersects $\overleftrightarrow{XY}$ .

So, $\overleftrightarrow{XY}$ is common internal tangent to both the circles.