Question

draw a circle of radius of 4 cm From a point 9 cm away from its centre draw two tangents to the circle measure the length of the tangent
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Answer 1

Solution :-

Given ,

• Radius of circle = 4cm
• Distance from the center to a point = 9cm

We need to find ,

• Length of tangent = ?.

As we know that ,

• L = √d² + r²

Where ,

• L is length of tangent .
• d is distance between the point and the centre.
• r is radius of the circle .

Substituting the known values we have ,

⇒ L = √9² + 4²

⇒ L = √81 + 16

⇒ L = √97

⇒ Length of tangent = 9.84 cm

Hence , length of tangent = 9.84 cm

Answer 2

Given:

• A circle of radius 4cm
• Two tangents from a point 9vm away from centre.

━━━━━━━━━━━━━━━━━━

Need to find:

• Length of the tangent =?

━━━━━━━━━━━━━━━━━━

Solution:

Steps to draw:

1. Draw a circle of radius 4 with centre O
2. Take a point 9cm away from the centre let that be P
3. Joint the point OP
4. Bisect the line OP, let the point of bisection be O'
5. Draw a circle taking O' as centre, let that circle intersect the previous circle at Q and R respectively.
6. Join PQ and PR.

PQ and PR are the required tangents!

PQ and PR are the required tangents!━━━━━━━━━━━━━━━━━━

━━━━━━━━━━━━━━━━━━

Now, Lets find the measure of the tangents!!

• Draw a Perpendicular from O to R!

Now :

• OR = Radius of circle = 4cm
• OP = 9cm (given)
• Angle ORP = 90° (by constriction)
• PR = tangent =?

So △ORP is a right angled triangle!

By Pythagoras theorem,

${PR}^{2} = {OR}^{2} + {OP}^{2}$

$\implies \: {PR}^{2} = {4}^{2} + {9}^{2} \\ \implies \: {PR}^{2} = 16 + 81 \\ \implies \: PR= \sqrt{97}$

$\red{\bold{\large{\boxed{\implies PR=9.8488cm}}}}$