TS EAMCET · Maths · Circle

The condition for the circles $x^{2} + y^{2} + a x + 4 = 0$ and $x^{2} + y^{2} + b y + 4 = 0$ to touch each other is

A $\frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{16}$
B $a^{2} + b^{2} = 16$
C $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{16}$
D $\frac{1}{a^{2}} + \frac{1}{b^{2}} = 4$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{16}$

Step-by-step Solution

Detailed explanation

S1x2+y2+ax+4=0 S2x2+y2+by+4=0 C1→(-a2,0), C2→(0, -b2) α1=a24-4, α2=b24-4 C1C2=α1+α2 a24+b24= a2-164+b2-164 ⇒a2+b2=a2-16+b2-16 Squaring a2+b2=a2+b2-32+2a2-16b2-16 16=a2-16b2-16 Squaring 256=(a2-16)(b2-16)…

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.

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The condition for the circles x2+y2+ax+4=0 and x2+y2+by+4=0 to touch each other is

Step-by-step Solution

The condition for the circles $x^{2} + y^{2} + a x + 4 = 0$ and $x^{2} + y^{2} + b y + 4 = 0$ to touch each other is