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The equations of the tangents drawn from the origin to the circle x^{2} + y^{2 }– 2rx – 2hy + h^{2} = 0, are (1988  2 Marks)
The given circle is x^{2} + y^{2} – 2rx – 2hy + h^{2} = 0 with centre (r, h) and radius = r.
Clearly circle touches yaxis so one of its tangent is x = 0.
Let y = mx be the other tangent through origin.
Then length of perpendicular from C (r, h) to y = mx should be equal to r.
⇒ m^{2}r^{2} – 2mrh + h^{2} = m^{2}r^{2} + r^{2}
∴ Other tangent is y =
or (h^{2} – r^{2}) x – 2rhy = 0
The number of common tangents to the circles x^{2} + y^{2} = 4 and x^{2} + y^{2} – 6x – 8y = 24 is (1998  2 Marks)
x^{2} + y^{2} = 4 (given) Centre C1 ≡ (0, 0) and R_{1} = 2.
Also for circle x^{2} + y^{2} – 6x – 8y – 24 = 0 C_{2} ≡ (3, 4) and R_{2} = 7.
Again C_{1} C_{2 }= 5 = R_{2} – R_{1}
Therefore, the given circles touch internally such that they can have just one common tangent at the point of contact.
If the circle x^{2} + y^{2} = a^{2} intersects the hyperbola xy = c^{2} in four points P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3}), S(x_{4}, y_{4}), then (1998  2 Marks)
Putting y = c^{2}/x in x^{2} + y^{2} = a^{2},
we obtain x^{2} + c^{4}/x^{2} = a^{2}
⇒ x^{4} – a^{2}x^{2} + c^{4} = 0 … (1)
As x_{1}, x_{2}, x_{3} and x_{4} are roots of (1),
⇒ x_{1} + x_{2 }+ x_{3} + x_{4} = 0 and x_{1} x_{2} x_{3} x_{4} = c_{4 }
Similarly, forming equation in y,
we get y_{1} + y_{2} + y_{3} + y_{4} = 0 and y_{1} y_{2} y_{3} y_{4} = c_{4}.
Circle(s) touching xaxis at a distance 3 from the origin and having an intercept of length on yaxis is (are) (JEE Adv. 2013)
There can be two possibilites for the given circle as shown in the figure
∴ The equations of circles can be (x – 3)^{2} + (y – 4)^{2} = 4^{2 }
or (x – 3)^{2 }+ (y + 4)^{2} = 4^{2 }
i.e. x^{2} + y^{2} – 6x – 8y + 9 = 0
or x^{2} + y^{2} – 6x + 8y + 9 = 0
A circle S passes through the point (0, 1) and is orthogonal to the circles (x – 1)^{2} + y^{2} = 16 and x^{2 }+ y^{2} = 1. Then (JEE Adv. 2014)
Let the equation of circle be x^{2 }+ y^{2} + 2gx + 2 f y + c = 0
It passes through (0, 1)
∴ 1 + 2f + c = 0 ...(i)
This circle is orthogonal to (x – 1)^{2} + y2 = 16
i.e. x^{2} + y^{2} – 2x – 15 = 0 and x^{2} + y^{2} – 1 = 0
∴ We should have 2g (– 1) + 2f (0) = c – 15 or 2g + c – 15 = 0 ...(ii)
and 2g(0) + 2f (0) = c – 1 or c = 1 ...(iii)
Solving (i), (ii) and (iii), we get c = 1, g = 7, f = – 1
∴ Required circle is x^{2} + y^{2} + 14x – 2y + 1 = 0
With centre (– 7, 1) and radius = 7
∴ (b) and (c) are correct options.
Let RS be the diameter of the circle x^{2} + y^{2} = 1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s) (JEE Adv. 2016)
Circle : x^{2} + y^{2 }= 1
Equation of tangent at P(cosθ, sinθ) x cosθ + y sinθ = 1 ...(1)
Equation of normal at P y = x tanθ ...(2)
Equation of tangent at S is x = 1
∴ Equation of line through Q and parallel to RS is
∴ Intersection point E of normal and
= x tanθ ⇒
∴ Locus of E or y^{2 }= 1 – 2x
It is satisfied by the points and
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