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Question: The length of tangent drawn from a point \(8cm\) away from the centre of the circle of radius \(6cm\...

The length of tangent drawn from a point 8cm8cm away from the centre of the circle of radius 6cm6cm is:
(A) 5cm\sqrt 5 cm
(B) 25cm2\sqrt 5 cm
(C) 5cm5cm
(D) 27cm2\sqrt 7 cm

Explanation

Solution

The given question revolves around the concepts of the circles and its properties. So, we are given a circle of radius 6cm6cm and a point that is 8cm8cm away from the centre of the circle. We know that the tangent to the circle forms a right angled triangle with the radius at the point of contact. So, a right angled triangle is formed in the figure.

Complete step-by-step solution:
So, the radius of the given circle is 6cm6cm.
Distance of point from the centre of the circle is 8cm8cm.
We know that the tangent to the circle forms a right angled triangle with the radius at the point of contact. So, a right angled triangle is formed. We can draw the figure for the given station as below:
Now, the radius of the circle is one of the legs of the right angled triangle and the line joining the centre of the circle with the external point is the hypotenuse of the right angled triangle.

Now, we have to find the length of tangent AB in the given figure.
So, we have to find the other leg of the right angled triangle OAB.
This can be done easily using the Pythagoras theorem.
Following the Pythagoras theorem, we have (Hypotenuse)2=(Base)2+(Altitude)2{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}.
So, applying the Pythagoras theorem in the given right angled triangle triangle ABC, we get,
(OB)2=(OA)2+(AB)2{\left( {OB} \right)^2} = {\left( {OA} \right)^2} + {\left( {AB} \right)^2}
(8cm)2=(6cm)2+(AB)2\Rightarrow {\left( {8cm} \right)^2} = {\left( {6cm} \right)^2} + {\left( {AB} \right)^2}
Shifting all the constants to the left side of the equation, we get,
64cm2=36cm2+(AB)2\Rightarrow 64c{m^2} = 36c{m^2} + {\left( {AB} \right)^2}
64cm236cm2=(AB)2\Rightarrow 64c{m^2} - 36c{m^2} = {\left( {AB} \right)^2}
(AB)2=28cm2\Rightarrow {\left( {AB} \right)^2} = 28c{m^2}
AB=27cm\Rightarrow AB = 2\sqrt 7 cm

So, the length of the tangent drawn from a point 8cm8cm away from the centre of the circle of radius 6cm6cm is 27cm2\sqrt 7 cm.

Note: For solving such type of question, where we need to find the third side of a triangle using the Pythagoras theorem, we need to know the position of right angle in the triangle beforehand since we need to know which of the three sides is the hypotenuse of the right angled triangle and then apply the Pythagoras theorem (Hypotenuse)2=(Base)2+(Altitude)2{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}. One must know some basic properties of the circles in order to solve such question.