(C) For the first circle $x^2 + y^2 - 10x + 16 = 0$,the center $C_1 = (5, 0)$ and radius $r_1 = \sqrt{5^2 - 16} = \sqrt{25 - 16} = \sqrt{9} = 3$.For t

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) For the first circle $x^2 + y^2 - 10x + 16 = 0$,the center $C_1 = (5, 0)$ and radius $r_1 = \sqrt{5^2 - 16} = \sqrt{25 - 16} = \sqrt{9} = 3$.For the second circle $x^2 + y^2 = r^2$,the center $C_2 = (0, 0)$ and radius $r_2 = r$.The distance between the centers $d = \sqrt{(5-0)^2 + (0-0)^2} = 5$.Two circles intersect at two distinct points if $|r_1 - r_2| Substituting the values,we get $|3 - r| From $5 2$.From $|3 - r| -2$.Combining these conditions,we get $2 < r < 8$.

Class 11 Mathematics 10-1.Circle and System of Circles Geometrical problems regarding circle and its properties

Difficult

The circles $x^2 + y^2 - 10x + 16 = 0$ and $x^2 + y^2 = r^2$ intersect each other in two distinct points,if

IIT 1994 All IIT

A
$r < 2$
B
$r > 8$
C
$2 < r < 8$
D
$2 \le r \le 8$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

10-1.Circle and System of Circles

Subtopic

Geometrical problems regarding circle and its properties

Previous Year

IIT 1994 Paper All IIT years →

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

EasyAP EAMCET 2004

View Solution

The angle subtended at the centre of a circle of radius $3 \text{ metres}$ by an arc of length $1 \text{ metre}$ is equal to

Easy

View Solution

If the two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect at two distinct points,then:

Difficult

View Solution

The area of the quadrilateral formed by the tangents from the point $(4,5)$ to the circle $x^2+y^2-4x-2y-11=0$,and the pair of radii joining the points of contact of these tangents is

DifficultAP EAMCET 2020

View Solution

The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The circles $x^2 + y^2 - 10x + 16 = 0$ and $x^2 + y^2 = r^2$ intersect each other in two distinct points,if

Similar Questions

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

The angle subtended at the centre of a circle of radius $3 \text{ metres}$ by an arc of length $1 \text{ metre}$ is equal to

If the two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect at two distinct points,then:

The area of the quadrilateral formed by the tangents from the point $(4,5)$ to the circle $x^2+y^2-4x-2y-11=0$,and the pair of radii joining the points of contact of these tangents is

The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module