The radical centre of the circles x2+y2-16 x+60=0, x2+y2-12 x+27=0, x2+y2-12 y+8=0 is

Question

The radical centre of the circles x2+y2-16 x+60=0, x2+y2-12 x+27=0, x2+y2-12 y+8=0 is (A) (13,33 / 4) (B) (33 / 4,-13) (C) (33 / 4,13) (D) None of these

Tardigrade Admin · Accepted Answer

S1 ≡ x2+y2-16 x+60=0 ...(i) S2 ≡ x2+y2-12 x+27=0 ...(ii) S3 ≡ x2+y2-12 y+8=0 ...(iii) The radical axis of circle (i) and circle (ii) is S1-S2=0 ⇒ -4 x+33=0 ... (iv) The radical axis of circle (ii) and circle (iii) is S2-S3=0 ⇒ -12+12 y+19=0 ldots( v ) Solving (iv) and (v), we get the radical centre ((33/4), (20/3))

Q. The radical centre of the circles $x^{2} + y^{2} - 16 x + 60 = 0$ ,
$x^{2} + y^{2} - 12 x + 27 = 0, x^{2} + y^{2} - 12 y + 8 = 0$ is

$(13, 33/4)$

$(33/4, - 13)$

$(33/4, 13)$

None of these

Q. The radical centre of the circles x2+y2−16x+60=0, x2+y2−12x+27=0,x2+y2−12y+8=0 is

(13,33/4)

(33/4,−13)

(33/4,13)

None of these

Q. The radical centre of the circles $x^{2} + y^{2} - 16 x + 60 = 0$ ,
$x^{2} + y^{2} - 12 x + 27 = 0, x^{2} + y^{2} - 12 y + 8 = 0$ is

$(13, 33/4)$

$(33/4, - 13)$

$(33/4, 13)$