વર્તુળ \({x^2} + {y^2} - 8x - 8y - 4 = 0\) ને બહારથી સ્પર્શતા તથા \( x-\) અક્ષને પણ સ્પર્શતા હોય તેવા વર્તૂળોના કેન્દ્રો . . . . . પર આવેલ છે.

A એક અતિવલય
B એક પરવલય
C એક વર્તૂળ
D એક ઉપવલય કે જે વર્તૂળ નથી.

Verified Solution

Answer & Solution

Correct Answer

(B) એક પરવલય

Step-by-step Solution

Detailed explanation

\(x^{2}+y^{2}-8 x-8 y-4=0\) Center \((4,4)\) Radius \(=\sqrt{4 z^{2}+4^{2}+4}=6\) Let centre of the circle is \((\mathrm{h}, \mathrm{k})\) \(\sqrt{(h-4)^{2}+(k-4)^{2}}=(6+k)\) \((h-4)^{2}+(k-4)^{2}=(6+k)^{2}\) \(h^{2}-8 h+16+k^{2}-8 k+16=36+k^{2}+12 k\) \(h^{2}-8 h-20 k-4=0\)…

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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