The centre of the smallest circle which cuts the circles x2+y2-2 x-4 y-4=0 and x2+y2-10 x+12 y+52=0 orthogonally is

Question

The centre of the smallest circle which cuts the circles x2+y2-2 x-4 y-4=0 and x2+y2-10 x+12 y+52=0 orthogonally is (A) (1,, 2) (B) (-3,2) (C) (3,-2) (D) (3,4)

Tardigrade Admin · Accepted Answer

Let the equation of the required circle be

x^{2} + y^{2} + 2 gx + 2 f y + c = 0

and centre is

(- g, - f)

Given circles,

C_{1} : x^{2} + y^{2} - 2 x - 4 y - 4 = 0

Here,

g_{1} = - 1, f_{1} = - 2, c_{1} = - 4

C_{2} : x^{2} + y^{2} - 10 x + 12 y + 52 = 0

Here,

g_{2} = - 5, f_{2} = 6, c_{2} = 52

Condition of two circles cuts orthogonally

2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}

For

C_{1} : 2 (g) (- 1) + 2 (f) (- 2) = C - 4

or

2 g + 4 f = - c + 4 \dots

(i)
For

C_{2} : 2 (g) (- 5) + 2 (f) (6) = c + 52

10 g - 12 f = - c - 52 \dots

.(ii)
Subtracting on Eqs. (i) and (ii), we get

- 8 g + 16 f = 56

or

g - 2 f = - 7

or

g = 2 f - 7 \dots

(iii)
Multiplying Eq. (i) by 5 and then subtracting Eq. (ii), we get

or

8 f = - c + 18

or

c = - 8 f + 18

Hence, radius

r = g^{2} + f^{2} - c

= (2 f - 7)^{2} + f^{2} (- 8 f + 18)

= 5 f^{2} - 20 f + 31

= 5 (f^{2} - 4 f) + 31 = 5 (f - 2)^{2} + 11

For minimum, value

f - 2 = 0

\Rightarrow f = 2

Putting the value of '

f

in Bq. (iii), we get

g = 2 \times 2 - 7 \Rightarrow g = - 3

Therefore, coordinate of centre is

(+ 3, - 2)

.

Q. The centre of the smallest circle which cuts the circles $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ and $x^{2} + y^{2} - 10 x + 12 y + 52 = 0$ orthogonally is

$(1,, 2)$

$(- 3, 2)$

$(3, - 2)$

$(3, 4)$

Q. The centre of the smallest circle which cuts the circles x2+y2−2x−4y−4=0 and x2+y2−10x+12y+52=0 orthogonally is

(1,,2)

(−3,2)

(3,−2)

(3,4)

Q. The centre of the smallest circle which cuts the circles $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ and $x^{2} + y^{2} - 10 x + 12 y + 52 = 0$ orthogonally is

$(1,, 2)$

$(- 3, 2)$

$(3, - 2)$

$(3, 4)$