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Question: Consider the circles $C_1: x^2 + y^2 = 16$ and $C_2: x^2 + y^2 - 12x + 32 = 0$. Which of the followi...

Consider the circles C1:x2+y2=16C_1: x^2 + y^2 = 16 and C2:x2+y212x+32=0C_2: x^2 + y^2 - 12x + 32 = 0. Which of the following statement is/are correct?

A

Number of common tangent to these circles is 3.

B

The point P with coordinates (4,1) lies outside the circle C1C_1 and inside the circle C2C_2.

C

Their direct common tangent intersect at (12,0).

D

Slope of their radical axis is not defined.

Answer

(A), (C), and (D)

Explanation

Solution

Circle C1C_1: x2+y2=16x^2 + y^2 = 16 has center O1=(0,0)O_1 = (0,0) and radius r1=4r_1 = 4.

Circle C2C_2: x2+y212x+32=0x^2 + y^2 - 12x + 32 = 0. Completing the square: (x6)2+y2=4(x-6)^2 + y^2 = 4. This circle has center O2=(6,0)O_2 = (6,0) and radius r2=2r_2 = 2.

The distance between the centers O1O_1 and O2O_2 is d=(60)2+(00)2=6d = \sqrt{(6-0)^2 + (0-0)^2} = 6.

The sum of the radii is r1+r2=4+2=6r_1 + r_2 = 4 + 2 = 6. Since d=r1+r2d = r_1 + r_2, the circles touch externally.

Let's analyze each statement:

(A) Number of common tangents: When two circles touch externally, they have exactly 3 common tangents (2 direct and 1 transverse). So, statement (A) is correct.

(B) Position of point P(4,1): For C1C_1: 42+1216=16+116=14^2 + 1^2 - 16 = 16 + 1 - 16 = 1. Since 1>01 > 0, P is outside C1C_1. For C2C_2: (46)2+124=(2)2+14=4+14=1(4-6)^2 + 1^2 - 4 = (-2)^2 + 1 - 4 = 4 + 1 - 4 = 1. Since 1>01 > 0, P is outside C2C_2. The statement claims P is outside C1C_1 (true) and inside C2C_2 (false). So, statement (B) is incorrect.

(C) Intersection of direct common tangents: The direct common tangents intersect at the exsimilicenter, which divides the line segment joining the centers externally in the ratio of their radii (r1:r2=4:2=2:1r_1:r_2 = 4:2 = 2:1). Using the external division formula for O1(0,0)O_1(0,0) and O2(6,0)O_2(6,0): T=(2×61×021,2×01×021)=(121,01)=(12,0)T = \left( \frac{2 \times 6 - 1 \times 0}{2-1}, \frac{2 \times 0 - 1 \times 0}{2-1} \right) = \left( \frac{12}{1}, \frac{0}{1} \right) = (12,0). So, statement (C) is correct.

(D) Slope of the radical axis: The radical axis is found by C1C2=0C_1 - C_2 = 0. (x2+y216)(x2+y212x+32)=0(x^2 + y^2 - 16) - (x^2 + y^2 - 12x + 32) = 0 12x48=012x - 48 = 0 x=4x = 4. This is a vertical line, whose slope is undefined. So, statement (D) is correct.

The correct statements are (A), (C), and (D).