The locus of the centre of the circle for which one end of the diameter is (3,3) while the other end lies on the line x+y=4.

3 years ago

Share

Answers : (1)

                                        

let (h,k) be the centre of the circle and other end of diameter be (x1,y1).


h= (x1+3)/2      and        k= (y1+3)/2


or, x1=2h-3       and        y1=2k-3


but (x1,y1) lie on x+y=4,


(2h-3)+(2k-3)=4


or, h+k=5


for locus, replace (h,k) by (x,y)..


thus, locus is  x+y=5


 

3 years ago

Post Your Answer

More Questions On Analytical Geometry

Ask Experts

Have any Question? Ask Experts
Post Question
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
Show that the perpendiculars falling from any point of the st.line 2x+11y=5 upon the two st. lines 24x+7y=20 and 4x-3y=2 are equal to each other.
 
 
solution: given L1 :24x+7y=20 ; slope M1 = -24/7 L2 ;4x-3y=2 ; slope M2= 4/3 and L3 :2x+11y=5 ; slope M3= -2/11 ifperpendiculars falling from any point from the line L3 to L1 and L2 areequal...
  img
Ajay Verma 6 months ago
Find the equation of circle passing through (3, -6) and touching both the axes ? (a) x 2 + y 2 – 6x + 6y +9 =0 (b) x 2 + y 2 + 6x – 6y +9 =0 (c) x 2 + y 2 – 30x + 30y + 225 =0 (d) x 2 + y 2...
 
 
solution: ans: “a” and “ c” both AS given circle is touching both the axis and passing through (3, -6) so its clear that circle would be in 4 th quadrant .. so the...
  img
Ajay Verma 6 months ago
Given two straight lines AB and AC whose equations are 3x + 4y = 5 and 4x – 3y = 15 respectively. Then the possible equation of line BC through (1, 2), such that ABC is isosceles, is L1 = 0...
 
 
Equation of the line BC passing through (1,2) with slope ‘m’ is (y-2) = m(x-2) Since ABC is isosceles, angle B = angle C. using Cos(theta) formula for finding angle between the lines AB,BC &...
 
Y RAJYALAKSHMI 6 months ago
At what values of 'a' do all the zeroes of the function, f(x) = (a-2)x 2 + 2ax + a + 3 lie on the interval (-2, 1)?
 
 
Hi, You need to consider 2 cases here. Case 1 : Take a-2>0 Now the concavity is upwards, draw the desired graph and think about the conditions. We need f(-2)>0 f(1)>0 \ But this is...
  img
Yash Baheti 4 months ago
 
The answer includes {2} . But leading coeff. is not equal to 0. Am i right ???
 
piyush soni 4 months ago
A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. Find the time after which it lands...
 
 
Hello Student, Well do this by relative motion. a_rel=-20 v_rel=10 0=10t-10t^2 =>t(10-10t)=0 =>t=0,1sec Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty
  img
Arun Kumar 6 months ago
a line passes through a point with position vector i(cap)-2j(cap)-k(cap) and is parallel to the vector i(cap)-2j(cap)+2k(cap) find the distance of point p(5,0,-4) from the line and how?
 
 
The equation of line with the given conditions : (x - 1)/1 = (y+2)/-2 = (z+1)/2 Now calculate the distance of this line from the point p by distance formula. Simple Thanks Bharat Bajaj...
  img
bharat bajaj 10 months ago
View all Questions »