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CAT 2020 Slot 2 — QA Question 8

2 CirclesEasy

Let C1 and C2 be concentric circles such that the diameter of C1 is 2 cm longer than that of C2. If a chord of C1 has length 6 cm and is a tangent to C2, then the diameter, in cm, of C1 is

Answer & solution

Answer: 10

Solution

Easy

The perpendicular from the common center to the chord of C1C_1 is exactly the radius of C2C_2 (since the chord is tangent to C2C_2) and it bisects the chord. This sets up a right triangle linking the two radii and half the chord.

O r r+1 3

Let radius of C2C_2 be rr; then radius of C1C_1 is r+1r+1 (diameters differ by 22). The 6 6\,cm chord of C1C_1 is tangent to C2C_2, so the center-to-chord distance is rr, and the foot of the perpendicular bisects the chord into halves of 3 3\,cm.

1

Set the radii. Diameter of C1C_1 exceeds C2C_2 by 22, so radii differ by 11.

rC2=r,rC1=r+1\begin{aligned} &r_{C_2} = r,\qquad r_{C_1} = r+1 \end{aligned}
2

Right triangle. Center OO, foot AA (where the radius meets the tangent chord), endpoint BB on C1C_1: OA=rOA=r, AB=3AB=3, OB=r+1OB=r+1.

OB2=OA2+AB2(Pythagoras)⇒ (r+1)2=r2+32\begin{aligned} &OB^2 = OA^2 + AB^2 \quad\text{(Pythagoras)}\\ &\Rightarrow\ (r+1)^2 = r^2 + 3^2 \end{aligned}
3

Solve. Expand and isolate rr.

r2+2r+1=r2+9⇒ 2r=8⇒ r=4\begin{aligned} &r^2 + 2r + 1 = r^2 + 9\\ &\Rightarrow\ 2r = 8\\ &\Rightarrow\ r = 4 \end{aligned}

Diameter of C1=2(r+1)=2×5=10 C_1 = 2(r+1) = 2\times 5 = 10\,cm.

Diameter of C1=10 cm\text{Diameter of } C_1 = 10\ \text{cm}
CAT 2020 Slot 2 QA Q8: Let C 1 and C 2 be concentric circles such that the diameter of C 1 is 2 cm longer than that of C 2 . If a cho — Solution | TheCATExam