JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 18)
A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :
an ellipse
a parabola
a hyperbola
a straight line
Explanation
Let equation of circle is
x2 + y2 + 2fx + 2fy + e = 0, it passes through (0, 2b)
$$ \Rightarrow $$ 0 + 4b2 + 2g $$ \times $$ 0 + 4f + c = 0
$$ \Rightarrow $$ 4b2 + 4f + c = 0 . . . (i)
$$2\sqrt {{g^2} - c} = 4a$$ . . . (ii)
g2 $$-$$ c = 4a2 $$ \Rightarrow $$ c = $$\left( {{g^2} - 4{a^2}} \right)$$
Putting in equation (1)
$$ \Rightarrow $$ 4b2 + 4f + g2 $$-$$ 4a2 = 0
$$ \Rightarrow $$ x2 + 4y + 4(b2 $$-$$ a2) = 0, it represent a hyperbola.
x2 + y2 + 2fx + 2fy + e = 0, it passes through (0, 2b)
$$ \Rightarrow $$ 0 + 4b2 + 2g $$ \times $$ 0 + 4f + c = 0
$$ \Rightarrow $$ 4b2 + 4f + c = 0 . . . (i)
$$2\sqrt {{g^2} - c} = 4a$$ . . . (ii)
g2 $$-$$ c = 4a2 $$ \Rightarrow $$ c = $$\left( {{g^2} - 4{a^2}} \right)$$
Putting in equation (1)
$$ \Rightarrow $$ 4b2 + 4f + g2 $$-$$ 4a2 = 0
$$ \Rightarrow $$ x2 + 4y + 4(b2 $$-$$ a2) = 0, it represent a hyperbola.
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