The length of the tangent drawn to a circle with radius $7 c m$ from a point $25 c m$ away from the centre is ___ $c m .$

Open in App

Solution

$∠ A B O = 90^{\circ}$ [point of contact]

Using Pythagoras theorem in $△ A B O$

$A O^{2} = A B^{2} + B O^{2}$

$\Rightarrow 25^{2} = A B^{2} + 7^{2}$

$\Rightarrow A B^{2} = 25^{2} - 7^{2}$

$\Rightarrow A B^{2} = 576$

$\Rightarrow A B = 24 c m$

Hence, length of the tangent $= 24 c m$

Suggest Corrections

114

Similar questions

A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is $x$ cm. The value of $x$ is ___ cm.

Q. Construct a circle of radius 7cm. From a point 10 cm away from its centre, draw tangents to the circle.

The length of the tangent drawn from a point $8 c m$ away from the centre of a circle of radius $6 c m$ is

Join BYJU'S Learning Program

The length of the tangent drawn to a circle with radius 7 cm from a point 25 cm away from the centre is ___ cm.

∠ABO=90∘ [point of contact] Using Pythagoras theorem in △ABO AO2=AB2+BO2 ⇒252=AB2+72 ⇒AB2=252−72 ⇒AB2=576 ⇒AB=24 cm Hence, length of the tangent =24 cm

The length of the tangent drawn to a circle with radius $7 c m$ from a point $25 c m$ away from the centre is ___ $c m .$

$∠ A B O = 90^{\circ}$ [point of contact]

Using Pythagoras theorem in $△ A B O$

$A O^{2} = A B^{2} + B O^{2}$

$\Rightarrow 25^{2} = A B^{2} + 7^{2}$

$\Rightarrow A B^{2} = 25^{2} - 7^{2}$

$\Rightarrow A B^{2} = 576$

$\Rightarrow A B = 24 c m$

Hence, length of the tangent $= 24 c m$