ACERISE INDIA
Construction Of Tangents To A Circle From A Point Outside The Circle
We know that a tangent is perpendicular to the radius through the point of contact. Therefore, if we want to draw a tangent at a point of a circle, we have to simply draw the radius through that point and draw a line perpendicular to that radius. It will be the required tangent at that point.
However, we do not follow this method, when a point lies outside the circle. This is because we do not know at which point on the circle the radius should be drawn so that the line joining that point and the point outside the circle is perpendicular to the radius.
Example 1: Draw a pair of tangents to a circle of radius 4 cm from a point outside the circle such that the tangents are inclined at an angle of 70°.
Solution:
We can draw the tangents using a property of circles. It is given that the tangents are inclined at an angle of 70°. Therefore, the radii joining the points of contacts to the centre of the circle are inclined at an angle of 110° as shown in the figure.
Firstly, let us draw a circle of radius 4 cm with the help of ruler and compasses. Let O be the centre of the circle.
We draw a radius OT of the circle. Now we draw another radius OQ making an angle of 110° with OT.
Now, we draw lines perpendicular to OT and OQ at points T and Q respectively. Let these two perpendicular lines intersect each other at a point P.
Thus, PT and PQ are the required tangents that are inclined at an angle of 70°.
Example 2: Draw a pair of tangents from point P to the given circle.
Solution:
In the given figure, the centre of the circle is not given. Therefore, first of all, we will find the centre of the circle.
For this, let us draw two non-parallel chords of the circle and draw their perpendicular bisectors. The point of intersection of these perpendicular bisectors is the centre O of the circle.
Now, we join the line segment OP and draw its perpendicular bisector. Let this perpendicular bisector intersect OP at point M.
Then we draw a circle with radius equal to OM taking M as centre. Let this circle intersect the given circle at points T and Q. We join PT and PQ.
Thus, PT and PQ are the required tangents from a point P outside the circle.
