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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

గమనిక : ప్రతిభ.ఈనాడు.నెట్లో వచ్చే ప్రకటనలు అనేక దేశాల నుండి, వ్యాపారస్తులు లేదా వ్యక్తుల నుండి వివిధ పద్ధతులలో సేకరించబడతాయి. ఆయా ప్రకటనకర్తల ఉత్పత్తులు లేదా సేవల గురించి ఈనాడు యాజమాన్యానికీ, ఉద్యోగస్తులకూ ఎటువంటి అవగాహనా ఉండదు. కొన్ని ప్రకటనలు పాఠకుల అభిరుచిననుసరించి కృత్రిమ మేధస్సు సాంకేతికతతో పంపబడతాయి. ఏ ప్రకటనని అయినా పాఠకులు తగినంత జాగ్రత్త వహించి, ఉత్పత్తులు లేదా సేవల గురించి తగిన విచారణ చేసి, తగిన జాగ్రత్తలు తీసుకొని కొనుగోలు చేయాలి. ఉత్పత్తులు / సేవలపై ఈనాడు యాజమాన్యానికి ఎటువంటి నియంత్రణ ఉండదు. కనుక ఉత్పత్తులు లేదా సేవల నాణ్యత లేదా లోపాల విషయంలో ఈనాడు యాజమాన్యం ఎటువంటి బాధ్యత వహించదు. ఈ విషయంలో ఎటువంటి ఉత్తర ప్రత్యుత్తరాలకీ తావు లేదు. ఫిర్యాదులు తీసుకోబడవు.

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