JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 22)
If a circle C, whose radius is 3, touches externally the circle,
$${x^2} + {y^2} + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
$${x^2} + {y^2} + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
$$2\sqrt 5 $$
$$3\sqrt 2 $$
$$\sqrt 5 $$
$$2\sqrt 3 $$
Explanation
Given circle is :
x2 + y2 + 2x $$-$$ 4y $$-$$4 = 0
$$\therefore\,\,\,$$ its center is ($$-$$ 1, 2) and radius is 3 units.
Let A = (x, y) be the center of the circle C
$$ \therefore $$$$\,\,\,$$ $${{x - 1} \over 2}$$ = 2 $$ \Rightarrow $$ x = 5 and $${{y + 2} \over 2}$$ = 2 $$ \Rightarrow $$ y = 2
So the center of C is (5, 2) and its radius is 3
$$\therefore\,\,\,$$ Equation of center C is :
x2 + y2 $$-$$ 10x $$-$$ 4y + 20 = 0
$$\therefore\,\,\,$$ The length of the intercept it cuts on the x-axis
= 2$$\sqrt {{g^2} - c} = 2\sqrt {25 - 20} = 2\sqrt 5 $$
x2 + y2 + 2x $$-$$ 4y $$-$$4 = 0
$$\therefore\,\,\,$$ its center is ($$-$$ 1, 2) and radius is 3 units.
Let A = (x, y) be the center of the circle C
$$ \therefore $$$$\,\,\,$$ $${{x - 1} \over 2}$$ = 2 $$ \Rightarrow $$ x = 5 and $${{y + 2} \over 2}$$ = 2 $$ \Rightarrow $$ y = 2
So the center of C is (5, 2) and its radius is 3
$$\therefore\,\,\,$$ Equation of center C is :
x2 + y2 $$-$$ 10x $$-$$ 4y + 20 = 0
$$\therefore\,\,\,$$ The length of the intercept it cuts on the x-axis
= 2$$\sqrt {{g^2} - c} = 2\sqrt {25 - 20} = 2\sqrt 5 $$
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