Question

equation of a circle S(x,y)=0 , (S(2,3)=16) which touches the line 3x+4y-7=0 at (1,1) is given by
A) x^2+y^2+2X+2Y-6=o
B)x^2+y^2+X+Y=0

Answer 1

solve the linear equation for x or y and sub into the circle equation. i.e x=(7-4y)/3 replace that as x in the circle and then it can be solved

Answer 2

What does (2,3)=16 mean?

Answer 3

when you put the point (2,3) in the equation of circle you get 16

Answer 4

Thank you ankit and taplin

Answer 5

@Ankitha007 have yous solved the question. I am sure how to proceed with Taplin's method

Answer 6

I am solving ... :) I've understood its A) answer I hope.

Answer 7

i will assume the eq of circle be (x-a)^2+(y-b)^2=r^2 (a,b,r) 3 unknowns
now we can make 3 equations
f(2,3)=16 f(1,1) =0 and b-1/a-1=4/3

Answer 8

can someone help me understand the method used by @Taplin44

Answer 9

if the line and circle intersect then the x and y values must be the same.

Answer 10

at that point anyway

Answer 11

@Taplin44 Yes at the point of contact values will be same... but not at all the points! I am not sure you can replace value from linear equation

Answer 12

And question says it touches at (1,1) not intersect so the line will be a tangent

Answer 13

Any which touches 3X+4y-7=o will be of the form S(x,y) = (x-1)^2 +(y-1)^2 + I ( 3X+4Y-7) = 0 u said S(x,y)=16 therefore I=1 therefore we get circle equation A)

Answer 14

@ankit042 uve understood ....

Answer 15

@Ankitha007 I guess I can recall this formula now...just thinking of the proof. The method I mentioned will work but this one will be faster!
Thanks

Answer 16

Yes just studied this but dint pass through my mind ... Thanx

Answer 17

1,1 is a point on the circle only equation a satisfies this condition xD

Answer 18

ya probably best to get familiar with this method though.. if you want to actually learn
"i will assume the eq of circle be (x-a)^2+(y-b)^2=r^2 (a,b,r) 3 unknowns
now we can make 3 equations
f(2,3)=16 f(1,1) =0 and b-1/a-1=4/3"

Answer 19

Yeah to understand yes