$X P$ and $X Q$ are tangents from $X$ to the circle with centre $O .$ $R$ is a point on the circle. Prove that, $X A + A R = X B + B R .$

Open in App

Solution

From the figure, there is an external point X from where two tangents, XP and XQ, are drawn to the circle.

$X P = X Q$

(The lengths of the tangents drawn from an external point to the circle are equal.)

Similarly,

$A P = A R$

$B Q = B R$

$X P = X A + A P$ --------- (1)

$X Q = X B + B Q$ --------- (2)

By substituting $A P = A R$ in equation (1) and $B Q = B R$ in equation $(2)$ , we get

$X P = X A + A R$

$X Q = X B + B R$

Since the tangents $X P$ and $X Q$ are equal, we get

$X A + A R = X B + B R .$

Suggest Corrections

Similar questions

In fig. XP and XQ are tangents from X to the circle with center O. R is a point on the circle. Prove that, XA + AR = XB + BR. [2 MARKS]

Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above, do the following:

In the fig., XP and XQ are tangents from T to the circle with centre O and R is any point on the circle. If AB is a tangent to the circle at R, prove that XA +AR = XB + BR

Q. In the fig.,

C P

and

C Q

are tangents from

C

to the circle with centre

O

and

R

is any point on the circle. If

A B

is a tangent to the circle at

R

, THEN

C A + A R = C B + B R .

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that :

$∠ P A Q = 2 ∠ O P Q$

TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Then prove that AP bisects $∠ T A B$ .

Join BYJU'S Learning Program

XP and XQ are tangents from X to the circle with centre O. R is a point on the circle. Prove that, XA+AR=XB+BR.

$X P$ and $X Q$ are tangents from $X$ to the circle with centre $O .$ $R$ is a point on the circle. Prove that, $X A + A R = X B + B R .$