All the tangents of a circle are perpendicular to the radius through the point of contact of that tangent.
OP is the radius of the circle and Q is any point on the line XY which is the tangent to the circle. As OP is the shortest line of all the distances of the point O to the points on XY. So OP is perpendicular to XY. Hence, OP XY
Find the radius of the circle, if the length of the tangent from point A which is 5 cm away from center is 4 cm.
As we know that the radius is perpendicular to the radius, so the ABO is a right angle triangle.
Given, AO = 5 cm and AB = 4 cm We can use Pythagoras theorem here
OA2 = OB2 + AB2
OB2 = OA2 – AB2
= 52 – 42
= 25 – 16
OB2 = 9
OB = 3 So the radius of the given circle is 3 cm
If two tangents PA and PB are drawn to a circle from a point P with centre O and OP is equal to the diameter of the circle then show that triangle APB is an equilateral triangle.
Given, is a tangent to the circle.
Therefore, (Tangent is perpendicular to the radius through the point of contact)
In (length of the tangents from the external point is equal)
(Angles opposite to equal sides are equal)
(Due to angle sum property)
Hence, is an equilateral triangle.