If the circle x2+y2+2 k x+4 y-4=0 has it's centre in 4 text th quadrant and touches the circle x2+y2+6 x-2 y+6=0, then k=

Question

If the circle x2+y2+2 k x+4 y-4=0 has it's centre in 4 text th quadrant and touches the circle x2+y2+6 x-2 y+6=0, then k= (A) -5 (B) (-15/7) (C) (-23/7) (D) -1

Tardigrade Admin · Accepted Answer

(-k,-2) arrow Centre of circle x2+y2+2 k x+4 y-4=0 since it lies in fourth quadrant -k >0 ⇒ k< 0 Centre of circle x2+y2+6 x-2 y+6=0 is (-3,1) and radius =√9+1-6 ∵ r1+r2=c1 c2 √k2+4+4+√9+1-6=√(k-3)2+(-2-1)2 √k2+8+2=√k2-6 k+18 k=-1 satisfies the equation.

Q. If the circle $x^{2} + y^{2} + 2 k x + 4 y - 4 = 0$ has it's centre in $4^{th}$ quadrant and touches the circle $x^{2} + y^{2} + 6 x - 2 y + 6 = 0$ , then $k =$

$- 5$

$\frac{- 15}{7}$

$\frac{- 23}{7}$

$- 1$

Q. If the circle x2+y2+2kx+4y−4=0 has it's centre in 4th quadrant and touches the circle x2+y2+6x−2y+6=0, then k=

−5

7−15​

7−23​

−1

(−k,−2)→ Centre of circle x2+y2+2kx+4y−4=0 since it lies in fourth quadrant −k>0⇒k<0 Centre of circle x2+y2+6x−2y+6=0 is (−3,1) and radius =9+1−6​ ∵r1​+r2​=c1​c2​ k2+4+4​+9+1−6​=(k−3)2+(−2−1)2​ k2+8​+2=k2−6k+18​ k=−1 satisfies the equation.

Q. If the circle $x^{2} + y^{2} + 2 k x + 4 y - 4 = 0$ has it's centre in $4^{th}$ quadrant and touches the circle $x^{2} + y^{2} + 6 x - 2 y + 6 = 0$ , then $k =$

$- 5$

$\frac{- 15}{7}$

$\frac{- 23}{7}$

$- 1$