In the given figure, PA and PB are two tangents to the circle from an external point P

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD

Given: PA and PB are the tangents to the circle.

PA = 12 cm

QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm (The lengths of tangents drawn from an external point to a circle are equal)

Similarly, QC = AC = 3 cm and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm

Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

Concept: Tangent to a Circle

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A number is chosen at random from the number –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?

S = {−3, −2, −1, 0, 1, 2, 3}Let E be the event of getting a number whose square is less than or equal to 1.So, E = {−1, 1, 0}P(E)=3/7.

Hence, the probability of getting a number whose square is less than or equal to is 3/7.

In figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Solution