Draw a circle of radius $3cm$ . Take two points $P$ and $Q$ on one of its extended diameters each at a distance of $7cm$ from its centre. Draw tangents to the circle from these two points $P$ and $Q$ . Give justification of the construction.
Answer
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Hint: In this question, we first need to construct a circle of radius $3cm$ , then take two points on any of its diameters and mark them on each side, namely, $P$ and $Q$ , at a distance of $7cm$ .
Then, we will find the perpendicular bisectors of the line segments $PO$ and $OQ$ , and draw a circle with the points where it intersects the line segments as centers. Now, these circles will cut the older circles at four points, mark them and join those points with $P$ and $Q$ .
Complete answer:
First, mark a point $O$ and draw a circle of radius $3cm$ , taking $O$ as the center.
Now, we will take any diameter of this circle and extend outwards by $7cm$ from the center from both directions.
Now, we have to draw tangents, from these two points to the circle, so for that, we draw perpendicular bisector of the line segment $PO$ . For that, first, open the compass mare than half of the line segment and mark an arc above and below by keeping the compass tip first on the point $P$ and then on the point $O$ . Then, join the two points $M$ and $N$ .
Perform, same steps for the line segment $QO$ , we get,
Now, taking $X$ and $Y$ as centers, draw two circles and mark the points where it cuts the older circle as $A,B,C,D$ , which gives,
Now, finally join the point $P$ with $A$ and $C$ and point $Q$ with $B$ and $D$ , the figure becomes,
Hence, $PA,PC,QB,QD$ are the required tangents.
Note:
Perpendicular bisector to a line segment is the perpendicular line which cuts the line segment into two halves.
Remember that we do not have to find the perpendicular bisector of the line segment $PQ$ , but we have to find the different perpendicular bisectors of the line segments $PO$ and $OQ$ .
In this question, the perpendicular bisector will not touch the older circle as half of $7cm$ i.e., $3.5cm$ is greater than the radius of the circle.
Then, we will find the perpendicular bisectors of the line segments $PO$ and $OQ$ , and draw a circle with the points where it intersects the line segments as centers. Now, these circles will cut the older circles at four points, mark them and join those points with $P$ and $Q$ .
Complete answer:
First, mark a point $O$ and draw a circle of radius $3cm$ , taking $O$ as the center.
Now, we will take any diameter of this circle and extend outwards by $7cm$ from the center from both directions.
Now, we have to draw tangents, from these two points to the circle, so for that, we draw perpendicular bisector of the line segment $PO$ . For that, first, open the compass mare than half of the line segment and mark an arc above and below by keeping the compass tip first on the point $P$ and then on the point $O$ . Then, join the two points $M$ and $N$ .
Perform, same steps for the line segment $QO$ , we get,
Now, taking $X$ and $Y$ as centers, draw two circles and mark the points where it cuts the older circle as $A,B,C,D$ , which gives,
Now, finally join the point $P$ with $A$ and $C$ and point $Q$ with $B$ and $D$ , the figure becomes,
Hence, $PA,PC,QB,QD$ are the required tangents.
Note:
Perpendicular bisector to a line segment is the perpendicular line which cuts the line segment into two halves.
Remember that we do not have to find the perpendicular bisector of the line segment $PQ$ , but we have to find the different perpendicular bisectors of the line segments $PO$ and $OQ$ .
In this question, the perpendicular bisector will not touch the older circle as half of $7cm$ i.e., $3.5cm$ is greater than the radius of the circle.
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