TS EAMCET · Maths · Circle

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, $k =$

A $- 5$
B $\frac{- 15}{7}$
C $\frac{- 23}{5}$
D $- 1$

Verified Solution

Answer & Solution

Correct Answer

(D) $- 1$

Step-by-step Solution

Detailed explanation

∙ x2+y2+2kx+4y-4=0 Centre C1=-k,-2 Radius R1=k2+4+4=k2+8 ∙ x2+y2+6x-2y+6=0 Centre C2=-3,1 Radius R2=9+1-6=2 ∙ Circle touch each others C1C2=R1+R2 ⇒-3+k2+1+22=k2+8+2 ⇒k2-6k+18=k2+8+2 ⇒k2-6k+18=k2+8+4+4k2+8 ⇒6-6k=4k2+8…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From TS EAMCET

Explore more questions on app

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=

Step-by-step Solution

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, $k =$