TheCATExam
Sign in

CAT 2020 Slot 2 — QA Question 9

Basics of CirclesEasy

Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

Answer & solution

  • 8.8

  • B

    6.6

  • C

    7.8

  • D

    7.2

Solution

Easy

Place OO at the origin. AA (north) and BB (east) are both inside the circle; the chord through them lies at a fixed distance from OO. Find ABAB, then the perpendicular distance from OO to line ABAB, then use the chord-length formula PQ=2R2−d2PQ = 2\sqrt{R^2 - d^2}.

O A B 4 3

Circle radius R=5R=5. AA is 4 4\,m north of OO, BB is 3 3\,m east of OO, so OA=4OA=4, OB=3OB=3 and ∠AOB=90∘\angle AOB=90^\circ. The chord PQPQ contains both AA and BB. Let dd be the perpendicular distance from OO to PQPQ and MM its foot.

1

Length ABAB. Right triangle AOBAOB with legs 33 and 44.

AB=32+42=5\begin{aligned} &AB = \sqrt{3^2 + 4^2} = 5 \end{aligned}
2

Distance from OO to the chord. OMOM is the altitude from the right angle of triangle AOBAOB to hypotenuse ABAB.

OM=OA⋅OBAB=4×35=2.4(altitude on hypotenuse)\begin{aligned} &OM = \frac{OA\cdot OB}{AB} = \frac{4\times 3}{5} = 2.4 \quad\text{(altitude on hypotenuse)} \end{aligned}
3

Chord length. With d=OM=2.4d=OM=2.4 and R=5R=5, use PQ=2R2−d2PQ = 2\sqrt{R^2-d^2}.

MP=R2−OM2=25−5.76=19.24⇒ PQ=2 MP=219.24≈2×4.39≈8.8\begin{aligned} &MP = \sqrt{R^2 - OM^2} = \sqrt{25 - 5.76} = \sqrt{19.24}\\ &\Rightarrow\ PQ = 2\,MP = 2\sqrt{19.24} \approx 2\times 4.39 \approx 8.8 \end{aligned}
PQ≈8.8 mPQ \approx 8.8\ \text{m}
CAT 2020 Slot 2 QA Q9: Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A a — Solution | TheCATExam