The shortest distance between a circle and an external point is r. The radius of the circle is also 'r'. What is the length of tangent from that point to the circle?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

The given problem is represented by the circle, Let the external point be P. Shortest distance between P

and circle is PQ where Q is the point of intersection between PO and circle.

Given that,

PQ=r.

Now in $△ O P R ∠ O R P = 90^{\circ}$

$∴ P R^{2} = r^{2} + (2 r)^{2}$

$P R = \sqrt{5 r^{2}} \Rightarrow \sqrt{5} r .$

Suggest Corrections

Similar questions

A tangent is drawn to a circle of radius r from an external point P .If length of tangent from a point to the circle is twice the minimum distance between circle and the point, what is the length of the tangent?

Q. The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is _______.

Q. If the radius of a circle is 5 cm & length of a tangent PA from an external point P is 12 cm.Find the distance of point P from the centre of circle.

Q. The length 'L' of a tangent, drawn from a point 'A' to a circle is

\frac{4}{3}

of the radius

r

. The shortest distance from A to the circle is

Q. Consider a circle with radius r. Tangents are drawn to it from an external point such that tangents are perpendicular to each other. All points from which such tangents can be drawn are connected by a smooth curve. M and N are any two points on that curve. What can be the maximum length of MN?

Join BYJU'S Learning Program

The shortest distance between a circle and an external point is r. The radius of the circle is also 'r'. What is the length of tangent from that point to the circle?

The correct option is D The given problem is represented by the circle, Let the external point be P. Shortest distance between P and circle is PQ where Q is the point of intersection between PO and circle. Given that, PQ=r. Now in △OPR∠ORP=90∘ ∴PR2=r2+(2r)2 PR=√5r2⇒√5r.