Question:
The locus of the centre of a circle, which touches externally the circle x2 + y2 - 6x - 6y + 14 = 0 and also touches the y-axis, is given by the equation
(a) x2 - 6x - 10y + 14 = 0
(b) x2 -10x - 6y + 14 = 0
(c) y2 - 6 x - 10y + 14 = 0
(d) y2 - 10x - 6y + 14 = 0
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Answers
Correct option d y2 - 10x - 6y + 14 = 0
Explanation :
Equation of AB is because AB is the chord of contact.
That is Therefore any circle passing through A and B is of the form
This represents semicircle of Δ PAB
⇒ it passes through P1 8. Therefore.
Therefore the semicircle of Δ PAB isx2 + y2 - 6x - 4y - 11 + 2x - 3y + 15 = 0
That is x2 + y2 - 4x - 10y + 19 = 0
Aliter: Suppose C3 2 is the centre of the circle.
Since ΔPAC = 90° = ΔPBC the semicircle of the quadrilateral PACB is also the semicircle of ΔPAB.
Therefore PC is a diameter of the semicircle of Δ PAB. Hence the equation of the semicircle is x - 1x - 3 + y - 8 y - 2 = 0.
That is y2 - 10x - 6y + 14 = 0
Answer:
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