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JEE • MATHEMATICS • SYSTEM OF CIRCLES

System of circles practice · Easy

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Showing 61-75 of 75 questions

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61 Easy
61 of 75
Analyzing +4 | -1
The two circles x2+y2−2x−2y−7=0{x^2} + {y^2} - 2x - 2y - 7 = 0 and 3(x2+y2)−8x+29y=03\left( {{x^2} + {y^2}} \right) - 8x + 29y = 0
View answer and explanation
Correct option: C
Cut each other orthogonally
Option C (Cut each other orthogonally) is correct because the condition 2(g1g2+f1f2)=c1+c22(g_1g_2 + f_1f_2) = c_1 + c_2 is satisfied, indicating that the circles intersect at right angles.
62 Easy
62 of 75
Application +4 | -1
The equation of the circle passing through (1,1)\left( {1,1} \right) and the points of intersection of x2+y2+13x−3y=0{x^2} + {y^2} + 13x - 3y = 0 and 2x2+2y2+4x−7y−25=02{x^2} + 2{y^2} + 4x - 7y - 25 = 0 is
View answer and explanation
Correct option: B
4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0
Option B (4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0) is correct because it represents the equation of the circle passing through the intersection points of the two given circles and the point (1,1).
63 Easy
63 of 75
Application +4 | -1
The straight line x+y−1=0x + y - 1 = 0 meets the circle x2+y2−6x−8y=0{x^2} + {y^2} - 6x - 8y = 0 at AA and BB. Then, the equation of the circle of which ABAB is a diameter is
View answer and explanation
Correct option: A
x2+y2−2y−6=0{x^2} + {y^2} - 2y - 6 = 0
Option A (x2+y2−2y−6=0{x^2} + {y^2} - 2y - 6 = 0) is correct because it represents a circle with AB as a diameter, derived from the intersection of the given line and circle.
64 Easy
64 of 75
Application +4 | -1
The value of λ\lambda, for which the circle x2+y2+2λx+6y+1=0{x^2} + {y^2} + 2\lambda x + 6y + 1 = 0 intersects the circle x2+y2+4x+2y=0{x^2} + {y^2} + 4x + 2y = 0 orthogonally, is
View answer and explanation
Correct option: C
−54\frac{{ - 5}}{4}
Option C (−54\frac{{ - 5}}{4}) is correct because it satisfies the orthogonality condition 2g1g2+2f1f2=c1+c22g_1g_2 + 2f_1f_2 = c_1 + c_2 for the given circles.
65 Easy
65 of 75
Application +4 | -1
If a circle passes through the point (a,b)(a, b) and cuts the circle x2+y2=4x^{2}+y^{2}=4 orthogonally, then the locus of its centre is
View answer and explanation
Correct option: B
2ax+2by−(a2+b2+4)=02 a x+2 b y-\left(a^{2}+b^{2}+4\right)=0
Option B (2ax+2by−(a2+b2+4)=02 a x+2 b y-\left(a^{2}+b^{2}+4\right)=0) is correct because it represents the locus of the center of a circle that passes through a given point and cuts another circle orthogonally.
66 Easy
66 of 75
Application +4 | -1
The radical centre of the circles
x2+y2−16x+60=0,x2+y2−12x+27=0\begin{array}{l} {x^2} + {y^2} - 16x + 60 = 0,\\ {x^2} + {y^2} - 12x + 27 = 0 \end{array}
and x2+y2−12x+8=0{x^2} + {y^2} - 12x + 8 = 0 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated radical center does not match any of the given options.
67 Easy
67 of 75
Application +4 | -1
The equation of circle described on the chord 3x+y+5=03 x+y+5=0 of the circle x2+y2=16x^{2}+y^{2}=16 as diameter is:
View answer and explanation
Correct option: A
x2+y2+3x+y−11=0x^{2}+y^{2}+3 x+y-11=0
Option A (x2+y2+3x+y−11=0x^{2}+y^{2}+3 x+y-11=0) is correct because it represents the equation of the circle formed with the given chord as diameter, found by using the family of circles equation.
68 Easy
68 of 75
Analyzing +4 | -1
The point (3,−4)\left( {3, - 4} \right) lies on both the circles x2+y2−2x+8y+13=0{x^2} + {y^2} - 2x + 8y + 13 = 0 and x2+y2−4x+6y+11=0{x^2} + {y^2} - 4x + 6y + 11 = 0. Then, the angle between the circles is
View answer and explanation
Correct option: D
450{45^0}
Option D (450{45^0}) is correct because it is the angle between the circles, calculated using the formula involving the distance between centers and the radii.
69 Easy
69 of 75
Application +4 | -1
The equation of a circle which cuts the three circles
x2+y2−2  x−6  y+14=0x2+y2−x−4  y+8=0x2+y2+2  x−6  y+9=0\begin{array}{l} {x^2} + {y^2} - 2\;x - 6\;y + 14 = 0\\ {x^2} + {y^2} - x - 4\;y + 8 = 0\\ {x^2} + {y^2} + 2\;x - 6\;y + 9 = 0 \end{array}
orthogonally, is
View answer and explanation
Correct option: A
x2+y2−2  x−4  y+1=0{x^2} + {y^2} - 2\;x - 4\;y + 1 = 0
The correct answer is A (x2+y2−2  x−4  y+1=0{x^2} + {y^2} - 2\;x - 4\;y + 1 = 0) because it represents the circle that cuts the three given circles orthogonally, determined by finding the radical center and the appropriate radius.
70 Easy
70 of 75
Application +4 | -1
Equation of the circle which cuts each of the circles x2+y2+2gx+c=0,x2+y2+2g1x+c=0x^2+y^2+2 g x+c=0, x^2+y^2+2 g_1 x+c=0 and x2+y2+2hx+2ky+a=0x^2+y^2+2 h x+2 k y+a=0 orthogonally is
View answer and explanation
Correct option: B
k(x2+y2)+(a−c)y−ck=0k\left(x^2+y^2\right)+(a-c) y-c k=0
Option B (k(x2+y2)+(a−c)y−ck=0k(x^2+y^2)+(a-c)y-ck=0) is correct because it represents the circle that cuts the given circles orthogonally, derived by setting up equations based on the orthogonality condition.
71 Easy
71 of 75
Application +4 | -1
Two equal circles with their centres on xx and yy axis will possess the radical axis in the following form
View answer and explanation
Correct option: B
2gx−2fy+g2−f2=02 g x-2 f y+g^2-f^2=0
Option B (2gx−2fy+g2−f2=02 g x-2 f y+g^2-f^2=0) is correct because it represents the general form of the radical axis when the centers lie on the x and y axes.
72 Easy
72 of 75
Application +4 | -1
The value of kk so that x2+y2+kx+4y+2=0x^2+y^2+k x+4 y+2=0 and 2(x2+y2)−4x−3y+k=02\left(x^2+y^2\right)-4 x-3 y+k=0 cut orthogonally, is
View answer and explanation
Correct option: C
−103\frac{-10}{3}
Option C (−103\frac{-10}{3}) is correct because the condition for orthogonal intersection of two circles (2(g1g2+f1f2)=c1+c22(g_1g_2 + f_1f_2) = c_1 + c_2) is applied to find the value of kk.
73 Easy
73 of 75
Application +4 | -1
A circle passes through the origin and has its centre on y=xy = x. If it cuts x2+y2−4  x−6  y+10=0{x^2} + {y^2} - 4\;x-6\;y + 10 = 0 orthogonally, then the equation of the circle is
View answer and explanation
Correct option: C
x2+y2−2  x−2  y=0{x^2} + {y^2} - 2\;x - 2\;y = 0
Option C x2+y2−2  x−2  y=0{x^2} + {y^2} - 2\;x - 2\;y = 0 is correct because it satisfies the condition of passing through the origin, having its center on y = x, and cutting the given circle orthogonally.
74 Easy
74 of 75
Understanding +4 | -1
The radical axis of two circles having centres at C1C_1 and C2C_2 and radii r1r_1 and r2r_2 is neither intersecting nor touching any of the circles, if
View answer and explanation
Correct option: B
0<C1C2<∣r1−r2∣0<\mathrm{C}_1 \mathrm{C}_2<\left|r_1-r_2\right|
Option B (0<C1C2<∣r1−r2∣0<\mathrm{C}_1 \mathrm{C}_2<\left|r_1-r_2\right|) is correct because it represents the condition where one circle is completely inside the other, and the radical axis lies outside both circles.
75 Easy
75 of 75
Understanding +4 | -1
The coaxial system of circles given by x2+y2+2gx+c=0{x^2} + {y^2} + 2gx + c = 0 for c<0c < 0 respects
View answer and explanation
Correct option: A
Intersecting circles
Option A (Intersecting circles) is correct because for c<0c < 0, the radical axis has two real points, indicating intersecting circles.

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