TS EAMCET · Maths · Circle

The equation of the tangent at the point $(0, 3)$ on the circle which cuts the circles $x^{2} + y^{2} - 2 x + 6 y = 0, x^{2} + y^{2} - 4 x - 2 y + 6 = 0$ and $x^{2} + y^{2} - 12 x + 2 y + 3 = 0$ orthogonally is

A $y = 3$
B $x = 0$
C $3 x + y - 3 = 0$
D $x + 3 y - 9 = 0$

Verified Solution

Answer & Solution

Correct Answer

(A) $y = 3$

Step-by-step Solution

Detailed explanation

Given equations of circles are S1≡x2+y2-2x+6y=0 S2≡x2+y2-4x-2y+6=0 S1≡x2+y2-12x+2y+3=0 Let the equation of circle which cuts the given circles orthogonally is x2+y2+2gx+2fy+c=0. From the condition of orthogonality: 2gg1+2ff1=c+c1 From S1⇒-2g+6f=c+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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The equation of the tangent at the point 0,3 on the circle which cuts the circles x2+y2-2x+6y=0, x2+y2-4x-2y+6=0 and x2+y2-12x+2y+3=0 orthogonally is

Step-by-step Solution

The equation of the tangent at the point $(0, 3)$ on the circle which cuts the circles $x^{2} + y^{2} - 2 x + 6 y = 0, x^{2} + y^{2} - 4 x - 2 y + 6 = 0$ and $x^{2} + y^{2} - 12 x + 2 y + 3 = 0$ orthogonally is