### CIRCLES (SOLVED PROBLEMS)

1. A point P moves in such a way that the ratio of its distances from two coplanar points is always a fixed number (≠ 1). Find the locus of the point.
Sol: Let two coplanar points be A(0, 0), B(a, 0)
Let P(x, y) be a point on the locus.
By the given data, = λ (constant) The locus of the point is a circle.

2. A triangle PQR inscribed in the circle x2 + y2 = 25. If 'Q' and 'R' have coordinates (3, 4) and (-4, 3) respectively, then find .
Sol:  We know that  O being the centre (0, 0) of
the given circle x2 + y2 = 25
Let m1 = slope of m2= slope of as m1m2 = -1;  3. If the abscissa and ordinates of two points 'P' and 'Q' are the roots of the equations
x2 + 2ax - b2 = 0 and x2 + 2px - q2 = 0 respectively. Find the equation of the circle with PQ as diameter.
Sol: Let x1, x2  and  y1, y2 be roots of x2 + 2ax − b2 = 0 and x2 + 2px - q2 = 0 respectively, then x1 + x2 = -2a; x1x2 = -b2
y1 + y2 = -2p; y1y2 = -q2
The equation of the circle with P(x1, y1), Q(x2, y2) as the end points of diameter is
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
x2 + y2 - x(x1 + x2) - y(y1 + y2) + x1x2 + y1y2 = 0
x2 + y2 + 2ax + 2py - b2 - q2 = 0.

4. Find the centre of the circle x = -1 + 2 cos ; y = 3 + 2 sin .
Sol: Given that = cos also = sin  (x + 1)2 + (y - 3)2 = 4 whose centre is (-1, 3).

5. Find the equation of the circle which touches both the axes and the straight line 4x + 3y = 6 in the first quadrant and lies below it.
Sol: Since the circle touches both the axes and the straight line 4x + 3y = 6 in first quadrant.
Then coordinates of its centre is (a, a) and radius = a where a > 0
Since 4x + 3y - 6 = 0 touches the circle Since (0, 0) and lies on the same side of the line 4x + 3y = 6 where (0, 0) and (3, 3) lie on the opposite side of the line.

Therefore for the required circle a = Hence equation of the required circle is 4x2 + 4y2 - 4x - 4y +1 = 0.

6. Find the equation of the circle which is touched by y = x has its centre on the positive direction of the X - axis and cuts off a chord of length '2' units along the line  y - x = 0.
Sol: Since the required circle has its centre on X - axis. So let the coordinates of the centre be (a, 0), the circle touches y = x.
Therefore radius = length of the perpendicular from (a, 0) on x - y = 0 is The circle cuts off a chord of length 2 units along x - y = 0.
From diagram, CP2 = CM2 + MP2 Thus centre of the circle is at (2, 0) and radius = So its equation is x2 + y2 - 4x + 2 = 0.

7. If a > 2b > 0, then find the +ve value of 'm' for which y = mx - b is a common tangent to x2 + y2 = b2 and (x - a)2 + y2 = b2.
Sol: y = mx - b ........ (1) is a tangent to the circle x2 + y2 = b2 for all values of m.
It is also tangent to the circle (x - a)2 + y2 = b2
radius = perpendicular distance from centre to (1) 8. Tangents are drawn to x2 + y2 = 1 from any arbitrary point 'P' on the line 2x + y − 4 = 0. The corresponding chord of contact passes through a fixed point whose coordinates are
Sol: Let any point on the line 2x + y − 4 = 0 be P ≡ (a, 4 − 2a). Equation of chord of contact of the circle
x+ y2 = 1 with respect to point P.
(x)(a) + y(4 − 2a) = 1
⇒ (4y − 1) + a(x − 2y) = 0
These lines pass through the point of intersection of 4y − 1 = 0 and x − 2y = 0 which is a fixed point whose co-ordinates are 9. Show that the circles x2 + y2 − 10x + 4y − 20 = 0 and x2 + y2 + 14x − 6y + 22 = 0 touch each other. Find the coordinates of the point of contact and the equation of the common tangent at the point of contact.
Sol: C1C2 = = r1 + r2
Hence the two circles touch externally
Let P be the point of contact of the two circles.
P divides the line segment C1C2 internally in the ratio 7 : 6  Since the two circles touch each other, then common tangent at the point of contact is S1 − S2 = 0
⇒ −24x + 10y − 42 = 0
⇒ 12x − 5y + 21 = 0 10. The common chord of x2 + y2 − 4x − 4y = 0 and x2 + y2 = 16 subtends at the origin an angle equal to
Sol: The equation of the common chord of the circles x2 + y2 − 4x − 4y = 0
and x2 + y2 = 16 is x + y = 4
which meets x2 + y= 16
at A(4, 0) and B(0, 4)
⇒ OA ⊥ OB
Hence the common chord AB makes a right angle at the centre of the circle x2 + y2 = 16.

11. If the circle x2 + y2 + 2x + 3y + 1 = 0 cuts x2 + y2 + 4x + 3y + 2 = 0 in A and B, then find the equation of the circle on AB as diameter

Sol: The equation of the common chord AB of two circles is 2x + 1 = 0
The equation of the required circle is
(x2 + y2 + 2x + 3y + 1 ) + λ (2x + 1) = 0 [Using S1 + λ (S1 - S2) = 0]
⇒ x2 + y2 + 2x (λ + 1) + 3y + λ + 1 = 0
Since AB is a diameter of this circle therefore centre lies on it.
So,  -2λ - 2 + 1 = 0 ⇒ λ = - 1/2
Thus the required circle is x2 + y2 + x + 3y + 1/2 = 0
or 2x2 + 2y2 + 2x + 6y + 1 = 0

12. Find the locus of the mid point of the chords of the circles x2 + y2 = a2 which subtend a right angle at the point (c, 0). Sol: Let P(h, k) be the midpoint of a chord BC with subtends a right angle at A (c, 0).
Then AP = PC = PB = Also = PC  From equations (i) and (ii) generalising (h, k), we get the locus of P is
(x - c)2 + y2 = a2 - (x2 + y2)
i.e.,  2(x2 + y2) - 2cx + c2 - a2 = 0

Posted Date : 19-02-2021

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గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.