The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units. The equation of the circle is
If a be the radius of a circle which touches xaxis at the origin, then its equation is
The equation of the circle with centre at (h,k) and radius equal to a is (x−h)^{2 }+(y−k)^{2 }= a^{2}
When the circle passes through the origin and centre lies on x− axis
⇒h = a and k = 0
Then the equation (x−h)^{2}+(y−k)^{2}=a^{2 }becomes (x−a)^{2}+y^{2}=a^{2}
If a circle passes through the origin and centre lies on x−axis then the abscissa will be equal to the radius of the circle and the y−coordinate of the centre will be zero Hence, the equation of the circle will be of the form
(x±a)^{2}+y^{2}=a^{2}⇒x^{2}+a^{2 }±2ax+y^{2}=a^{2}
=x^{2} +y^{2 }±2ax=0 is the required equation of the circle.
The equation of the circle passing through (3, 6) and whose centre is (2, –1) is
The equation of a circle which passes through the three points (3, 0) (1, –6), (4, –1) is
Gen equation of a circle is (x−h)^{2}+(y−k)^{2}=r^{2} (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)^{2}+k^{2}=r^{2} ___ (I)
(1−h)^{2} +(−6−k)^{2} = r^{2}(1−h)^{2}+(6+k)^{2}=r^{2} ___ (II)
From (I) & (II)
(3−h)^{2}+k^{2} =(1−h)^{2}+(6+k)^{2}
9+h^{2}−6h+k^{2} =1+h^{2} 2h+36+k^{2}+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)^{2}+(−1−k)^{2}=r^{2} ___ (IV)
From (I) & (IV)
(3−h)^{2}+k^{2} = (4−h)^{2} +(1+k)^{2}
9+h^{2} −6h+k^{2}
=16+h^{2} −8h+1+k^{2}+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)^{2} +(y+11/4)^{2} =170/16
x^{2}+25/16−10x/4+y^{2} +121/16+22y/4=170/16
Simplifying, we get 2x^{2}+xy^{2}−5x+11y−3=0
y = √3 + c_{1} & y = √3 + c_{2} are two parallel tangents of a circle of radius 2 units, then c_{1} – c_{2} is equal to
For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = c_{1}−c_{2}/(1+3)^{1/2}
⇒∣c_{1}−c_{2}∣ = 8
B and C are fixed point having coordinates (3, 0) and (–3, 0) respectively. If the vertical angle BAC is 90º, then the locus of the centroid of the DABC has the equation
Let A (a, b) and G (x. y) Now A, G, O are collinear
⇒ x = (2*0 + a)/3
⇒ a = 3x
and similarly b = 3y
Now (a,b) lies on the circle = 9x^{2} + 9y^{2} = 9
=> x^{2} + y^{2} = 1
The area of an equilateral triangle inscribed in the circle x^{2} + y^{2} – 2x = 0 is
The length of intercept on yaxis, by a circle whose diameter is the line joining the points (–4,3) and (12,–1) is
Equation of circle is given by,
The gradient of the tangent line at the point (a cos a, a sin a) to the circle x^{2} + y^{2} = a^{2}, is
lx + my + n = 0 is a tangent line to the circle x^{2} + y^{2} = r^{2}, if
If y = c is a tangent to the circle x^{2}+y^{2}–2x+2y–2 = 0 at (1, 1), then the value of c is
For line y=c to be tangent to the given circle at point (1,1)
It has to pass through (1,1)
⇒ c = 1
Line 3x + 4y = 25 touches the circle x^{2} + y^{2} = 25 at the point
Equation of tangent at (x1 , y1) to the circle x^{2} + y^{2} =25 is given by,
xx1 + yy1 = 25. Now comparing it with 3x+4y = 25
We get required point of contact (3,4)
The equations of the tangents drawn from the point (0, 1) to the circle x^{2} + y^{2}  2x + 4y = 0 are
Let equation of tangent with slope =m and point (0,1)
(y−1)=m(x−0)⇒y=mx+1
Intersection point
x^{2}+(mx+1)^{2}−2x+4(mx+1)=0
(1+m^{2})x^{2}+(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m−2)^{2}−4×5(1+m^{2})=0
36m^{2}+4−24m−20m^{2}+20=0
16m^{2}−20m+24=0
⇒ 2m^{2}−3m−2=0
(2m+1)(m−2)=0
The greatest distance of the point P(10, 7) from the circle x^{2} + y^{2} – 4x – 2y – 20 = 0 is
The equation of the normal to the circle x^{2}+y^{2} = 9 at the point is
(xx1)/(x1+g) = (yy1)/(y1+g)
[x1/(2)^{½}]/(1/(2)^{½}) = [y1/(2)^{½}]/(1/(2)^{½})
[(2x)^{½} 1 *(2)^{½}]/(2)^{½} = [(2x)^{½} 1 *(2)^{½}]/(2)^{½}
(2x)^{½} 1 = (2y)^{½} 1
= xy = 0
The parametric coordinates of any point on the circle x^{2} + y^{2} – 4x – 4y = 0 are
The length of the tangent drawn from the point (2, 3) to the circles 2(x^{2} + y^{2}) – 7x + 9y – 11 = 0.
Tangents are drawn from (4, 4) to the circle x^{2} + y^{2} – 2x – 2y – 7 = 0 to meet the circle at A and B. The length of the chord AB is
The angle between the two tangents from the origin to the circle (x – 7)^{2} + (y + 1)^{2} = 25 equals
Let tangent from origin be y = mx
Using the condition of tangency, we get
The angle between tangents = π/2
Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x^{2}+ y^{2} = 4. Their variable chord of contact always passes through a fixed point whose coordinates are
Let P(x_{1},y_{1}) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x_{1},y1_{)} w.r.t circle x2 + y2 = 4
xx_{1} + yy_{1} − 4 = 0 ...(1)
Also 3x_{1} + 4y1 − 12 = 0
⇒ x_{1} + 4/3y_{1} − 4 = 0 ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)
The locus of the midpoints of the chords of the circle x^{2} + y^{2} – 2x – 4y – 11 = 0 which subtend 60º at the centre is
Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a rightangled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)^{2}+(y−2)^{2}=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)^{½}]/2
= 2(3)^{1/2} units
Therefore, PC=2(3)^{1/2}
⇒ PC^{2}=12
⇒ (x−1)^{2}+(y−2)^{2}=12
⇒ x^{2}+y^{2}−2x−4y+5=12
⇒ x^{2}+y^{2}−2x−4y−7=0
The locus of the centres of the circles such that the point (2, 3) is the mid point of the chord 5x + 2y = 16 is
Slope of the given chord = −5/2
Slope of the line joining the midpoint on the chord and the centre of the circle = (3+f)/(2+g)
(5/2)[(3+f)/(2+g)] = −1
⇒ 15 + 5f = 4 + 2g
⇒ Locus is 2x − 5y + 11 = 0
The equation of the circle having the lines y^{2} – 2y + 4x – 2xy = 0 as its normals & passing through the point (2, 1) is
The normal line to circle is →y²  2 y + 4 x 2 xy=0
→ y(y2)  2x(y2)=0
→ (y2)(y2x)=0
the two lines are , y=2 and 2 x y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x y=0, we get
→2 x 2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(12)² + (21)²]^{½ }
=(1+1)^{½} =(2)^{½}
The equation of circle having center (1,2) and radius √2 is
= (x1)²+(y2)²=[√2]²
→ (x1)²+(y2)²= 2
x²+y² 2x4y+3 = 0
A circle is drawn touching the xaxis and centre at the point which is the reflection of (a, b) in the line y – x = 0. The equation of the circle is
Radius of the circle = a
and centre ≡ (b,a)
And circle is touching the xaxis.
The equation of the circle :
x^{2}+y^{2}−2bx−2ay+b^{2 }= 0
The number of common tangents of the circles x^{2} + y^{2} – 2x – 1 = 0 and x^{2} + y^{2} – 2y – 7 = 0
Internally touch ∴ common tangent is one.
The point from which the tangents to the circles x^{2} + y^{2} – 8x + 40 = 0, 5x^{2} + 5y^{2} – 25 x + 80 = 0, x^{2} + y^{2} – 8x + 16y + 160 = 0 are equal in length is
The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x  24 = 0,16y + 120 = 0 and  3x + 16y + 80 = 0. On solving, the required point is (8, 15/2).
If the circle x^{2} + y^{2} = 9 touches the circle x^{2} + y^{2} + 6y + c = 0, then c is equal to
touches the another circle
Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
And radius,
As both touches each other
So,
The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y –1 = 0 to the circle x^{2} + y^{2} + 2x – 4y = 0 is
The length of the common chord of circles x^{2} + y^{2} – 6x – 16 = 0 and x^{2} + y^{2} – 8y – 9 = 0 is
The distance between the chords of contact of tangents to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 from the origin and from the point (g, f) is
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