Skip to main content
Back to CIRCLES

Your Progress

0 / 480 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • CIRCLES

Circles practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 271-300 of 480 questions

Preparing practice and saved progress…

271 Easy
271 of 480
Application +4 | -1
The locus of the centre of the circle for which one end of a diameter is (1,1)\left( {1,1} \right) while the other end is on the line x+y=3x + y = 3, is
View answer and explanation
Correct option: C
2x+2y=52x + 2y = 5
Option C (2x+2y=52x + 2y = 5) is correct because it represents the locus of the center of the circle derived from the given conditions.
272 Easy
272 of 480
Application +4 | -1
If the lines 2x−3y−5=02 x-3 y-5=0 and 3x−4y=73 x-4 y=7 are diameters of a circle of area 154 square units, then the equation of the circle is
View answer and explanation
Correct option: C
x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0
Option C (x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0) is correct because the center and radius are correctly calculated from the given information.
273 Easy
273 of 480
Application +4 | -1
The equation of the tangents to the circle x2+y2=4{x^2} + {y^2} = 4, which are parallel to x+2y+3=0x + 2y + 3 = 0, are
View answer and explanation
Correct option: C
x+2y=±25x + 2y = \pm 2\sqrt 5
Option C (x+2y=±25x + 2y = \pm 2\sqrt 5) is correct because these are the equations of the tangents parallel to the given line, found by equating the perpendicular distance from the center to the radius.
274 Easy
274 of 480
Application +4 | -1
The length of the longest ray drawn from the point (4,3)(4,3) to the circle x2+y2+16x+18y+1=0x^2+y^2+16 x+18 y+1=0 is equal to
View answer and explanation
Correct option: A
the radius of the circle
Option A (the radius of the circle) is correct because the longest ray from a point to a circle is along the line connecting the point to the center of the circle, extending to the circle's circumference. The length of the tangent is equal to the radius in this case.
275 Easy
275 of 480
Application +4 | -1
If the square of the length of the tangent drawn from any point on the circle x2+y2+70x−45y+6789=0x^2+y^2+70 x-45 y+6789=0 to the circle x2+y2+70x−45y+9876=0x^2+y^2+70 x-45 y+9876=0 is KK then find K−3080.K-3080.
View answer and explanation
Correct option: A
7
Option A (7) is correct because the square of the tangent length is found to be 3087, and 3087 - 3080 = 7.
276 Easy
276 of 480
Application +4 | -1
The radius of the circle r2−22r(cos⁡θ+sin⁡θ)−5=0,{r^2} - 2\sqrt 2 r\left( {\cos \theta + \sin \theta } \right) - 5 = 0, is
View answer and explanation
Correct option: C
3
Option C (3) is correct because converting the polar equation to Cartesian form allows for easy identification of the circle's radius using the standard equation.
277 Easy
277 of 480
Application +4 | -1
Find the equation to the circle touching the yy-axis at a distance -3 from the origin and intercepting a length 8 on the xx-axis.
View answer and explanation
Correct option: C
x2+y2±10x+6y+9=0x^2+y^2 \pm 10 x+6 y+9=0
Option C (x2+y2±10x+6y+9=0x^2+y^2 \pm 10 x+6 y+9=0) is correct because it satisfies the conditions of touching the y-axis at (0, -3) and intercepting a length of 8 on the x-axis.
278 Easy
278 of 480
Analyzing +4 | -1
α,β\alpha, \beta and γ\gamma are parametric angles of three points P,QP, Q and RR respectively, on the circle x2+y2=1x^2+y^2=1 and AA is the point (−1,0)(-1,0). If the lengths of the chords AP,AQA P, A Q and ARA R are in GPG P, then cos⁡(α2),cos⁡(β2)\cos \left(\frac{\alpha}{2}\right), \cos \left(\frac{\beta}{2}\right) and cos⁡(γ2)\cos \left(\frac{\gamma}{2}\right) are in
View answer and explanation
Correct option: B
GPG P
Option B (GP) is correct because if the lengths of the chords AP, AQ, and AR are in GP, then the cosines of half the angles are also in GP.
279 Easy
279 of 480
Application +4 | -1
If the points (2,0),(0,1),(4,5)\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right) and (0,c)\left( {0,c} \right) are concyclic, then the value of cc is
View answer and explanation
Correct option: B
143\frac{{14}}{3}
Option B (14/3) is correct because it's the value of c that satisfies the concyclic condition after finding the circle's equation.
280 Easy
280 of 480
Analyzing +4 | -1
Let PQP Q and RSR S be tangents at the extremities of diameter PRP R of a circle of radius rr. If PSP S and RQR Q intersect at a point XX on the circumference of the circle, then 2r2 r equals
View answer and explanation
Correct option: A
PQ⋅RS\sqrt{P Q \cdot R S}
Option A (PQ⋅RS\sqrt{P Q \cdot R S}) is correct because using similar triangles formed by the intersecting tangents and the diameter, we can show that 2r=PQ⋅RS2r = \sqrt{PQ \cdot RS}.
281 Easy
281 of 480
Application +4 | -1
The radius of any circle touching the lines 3x−4y+5=03x - 4y + 5 = 0 and 6x−8y−9=06x - 8y - 9 = 0 is
View answer and explanation
Correct option: B
0.950.95
Option B (0.95) is correct because the radius of the circle touching two parallel lines is half the distance between the lines, which is calculated to be 0.95.
282 Easy
282 of 480
Application +4 | -1
Let SS be the circle in the xyx y-plane defined by the equation x2+y2=4x^2+y^2=4. Let PP be a point on the circle with both coordinates being positive. Let the tangent to SS at PP intersect the coordinate axes at MM and NN. Then the mid-point of the line segment MN must lie on the curve
View answer and explanation
Correct option: D
x2+y2=x2y2x^2+y^2=x^2 y^2
Option D (x+22y=6x+2 \sqrt{2} y=6) is correct because it is a common tangent to both circles.
283 Easy
283 of 480
Application +4 | -1
Radius of bigger circle touching the circle x2+y2−4x−4y+4=0x^2+y^2-4 x-4 y+4=0 and both the coordinate axes, is
View answer and explanation
Correct option: B
2(3+22)2(3+2 \sqrt{2}) units
Option B (2(3+22)2(3+2 \sqrt{2}) units) is correct because it represents the radius of the larger circle touching the given circle and both coordinate axes, derived from geometric considerations and solving for the unknown radius.
284 Easy
284 of 480
Application +4 | -1
The angle between the tangents drawn at the points (5,12)\left( {5,12} \right) and (12,−5)\left( {12, - 5} \right) to the circle x2+y2=169{x^2} + {y^2} = 169 is
View answer and explanation
Correct option: D
900{90^0}
Option D (900{90^0}) is correct because the tangents at the given points are perpendicular to each other, which can be verified by checking the product of their slopes.
285 Easy
285 of 480
Application +4 | -1
The values of λ\lambda so that the line 3x−4y=λ3x - 4y = \lambda touches x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0 are
View answer and explanation
Correct option: A
−35,  15- 35,\;15
Option A (−35,15-35, 15) is correct because these are the values of λ\lambda for which the line is tangent to the circle, found by equating the perpendicular distance from the center to the radius.
286 Easy
286 of 480
Analyzing +4 | -1
Let the tangent to the circle C1:x2+y2=2C_1: x^2+y^2=2 at the point M(−1,1)M(-1,1) intersect the circle C2:(x−3)2+(y−C_2:(x-3)^2+(y- 2)2=52)^2=5, at two distinct points AA and BB. If the tangents to C2C_2 at the points AA and BB intersect at NN, then the area of the triangle ANBA N B is equal to
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because the area of the triangle ANB is found by determining the intersection points A and B, finding the intersection point N of the tangents, and then calculating the area using determinant formula.
287 Easy
287 of 480
Application +4 | -1
Assertion: Radius of the circle 2x2+2y2+3x+4y+98=02 x^2+2 y^2+3 x+4 y+\frac{9}{8}=0 is 1.

Reason: Radius of the circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 is (12 coeff. of x)2+(12 coeff. of y)2− constant term \sqrt{\left(\frac{1}{2} \text { coeff. of } x\right)^2+\left(\frac{1}{2} \text { coeff. of } y\right)^2-\text { constant term }}
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because both the assertion about the radius and the reason explaining the radius formula are accurate and related.
288 Easy
288 of 480
Analyzing +4 | -1
Let A(1, 4)A(1,\,4) and B(1, −5)B\left( {1,\, - 5} \right) be two points. Let PP be a point on the circle (x−1)2+(y−1)2=1{(x - 1)^2} + {(y - 1)^2} = 1, such that (PA)2+(PB)2{\left( {PA} \right)^2} + {\left( {PB} \right)^2} have maximum value, then the points PP, AA and BB lie on
View answer and explanation
Correct option: A
a straight line
Option A (a straight line) is correct because the points P, A, and B are collinear when (PA)2 + (PB)2 is maximized, resulting in a straight line.
289 Easy
289 of 480
Application +4 | -1
The equation of circle is x2+y2−2x=0{x^2} + {y^2} - 2x = 0. The point P(−1,0)P\left( { - 1,0} \right) lies
View answer and explanation
Correct option: C
Outside the circle
Option C (Outside the circle) is correct because substituting the point P(-1, 0) into the circle's equation results in a positive value, indicating the point lies outside the circle.
290 Easy
290 of 480
Analyzing +4 | -1
If (a,0)(a, 0) is a point on a diameter of the circle x2+y2=4x^2+y^2=4, then the equation x2−4x−a2=0x^2-4 x-a^2=0 has
View answer and explanation
Correct option: D
all of these
Option D (all of these) is correct because the given conditions lead to the equation having a root in [-1, 0] and [2, 5], and distinct roots greater than -1 and less than 5.
291 Easy
291 of 480
Application +4 | -1
Equation of the circle with centre on the yy-axis and passing through the origin and (2,3) is
View answer and explanation
Correct option: B
3x2+3y2−13y=03{x^2} + 3{y^2} - 13y = 0
Option B (3x2+3y2−13y=03{x^2} + 3{y^2} - 13y = 0) is correct because it is derived from the conditions that the circle's center lies on the y-axis and passes through the origin and the point (2,3).
292 Easy
292 of 480
Application +4 | -1
The locus of the point, such that tangents drawn from it to the circle x2+y2−6x−8y=0x^2+y^2-6 x-8 y=0 are perpendicular to each other, is
View answer and explanation
Correct option: A
x2+y2−6x−8y−25=0x^2+y^2-6 x-8 y-25=0
Option A, x2+y2−6x−8y−25=0x^2+y^2-6x-8y-25=0, is the correct locus because it represents the director circle of the given circle, where tangents from any point on it are perpendicular.
293 Easy
293 of 480
Application +4 | -1
If circles x2+y2+2ax+c2=0x^2+y^2+2 a x+c^2=0 and x2+y2+2by+c2=0x^2+y^2+2 b y+c^2=0, touch each other then
View answer and explanation
Correct option: B
1a2+1b2=1c2\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}
Option B (1a2+1b2=1c2\frac{{1}}{{a^2}}+\frac{{1}}{{b^2}}=\frac{{1}}{{c^2}}) is correct because it is derived from the condition that the distance between the centers equals the sum or difference of the radii for touching circles.
294 Easy
294 of 480
Application +4 | -1
Tangents PAP A and PBP B are drawn to the circle x2+y2=8x^2+y^2=8 from any arbitrary point PP on the line x+y=4x+y=4. The locus of the mid-point of chord of contact ABA B is
View answer and explanation
Correct option: B
x2+y2−2x−2y=0x^2+y^2-2 x-2 y=0
Option A (x−3y=1x-\sqrt{3} y=1) is correct because it is a tangent to the circle (x−3)2+y2=1(x-3)^2+y^2=1 and is perpendicular to the tangent at P.
295 Easy
295 of 480
Application +4 | -1
Let the circle C1:x2+y2−2(x+y)+1=0C_1: x^2+y^2-2(x+y)+1=0 and C2C_2 be a circle having centre at (−1,0)(-1,0) and radius 2. If the line of the common chord of C1{C}_1 and C2{C}_2 intersects the y\mathrm{y}-axis at the point P,{P}, then the square of the distance of P{P} from the centre of C1{C}_1 is:
View answer and explanation
Correct option: A
2
Option A (2) is correct because the square of the distance is calculated using the coordinates of point P and the center of circle C1.
296 Easy
296 of 480
Understanding +4 | -1
The equation (x−x1)(x−x2)+(y−y1)(y−y2)=0\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0 represents a circle whose centre is
View answer and explanation
Correct option: B
(x1+x22,y1+y22)\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)
Option B ((x1+x22,y1+y22)\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)) is correct because the center of the circle is the midpoint of the diameter, and the given equation represents a circle with endpoints of a diameter at (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2).
297 Easy
297 of 480
Application +4 | -1
If xα+yβ=1\frac{x}{\alpha}+\frac{y}{\beta}=1 touches the circle x2+y2=a2x^2+y^2=a^2, then point (1α,1β)\left(\frac{1}{\alpha}, \frac{1}{\beta}\right) lies on
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the condition for tangency leads to the equation 1α2+1β2=1a2\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{1}{a^2}, which represents a circle.
298 Easy
298 of 480
Application +4 | -1
If A={(x,y):x2+y2=25}A=\left\{(x, y): x^2+y^2=25\right\} and B={(x,y):x2+9y2=144}B=\left\{(x, y): x^2+9 y^2=144\right\} then the number of points, A∩B{A} \cap {B} contains is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the intersection of a circle and an ellipse can have a maximum of 4 points.
299 Easy
299 of 480
Application +4 | -1
A\mathrm{A} line is drawn through a fixed point P(α,β)P(\alpha, \beta) to cut the circle x2+y2=r2x^2+y^2=r^2 at AA and B. Then PA.PB is equal to
View answer and explanation
Correct option: B
α2+β2−r2\alpha^2+\beta^2-r^2
Option B (α2+β2−r2\alpha^2+\beta^2-r^2) is correct because it represents the power of the point P with respect to the circle, which is equal to PA.PB.
300 Easy
300 of 480
Application +4 | -1
The line x+ytan⁡θ=cos⁡θx+y \tan \theta=\cos \theta touches the circle x2+y2=4x^2+y^2=4 for
View answer and explanation
Correct option: D
no value of θ\theta
Option D (no value of θ\theta) is correct because the condition for the line to touch the circle leads to an impossible value for cos⁡θ\cos \theta.
Revision notes and key takeaways

Review CIRCLES in your textbook or class notes, then use this set to check your understanding.

Showing 271-300 of 480 questions

Your practice

0 / 480 answered · 480 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.