Question

Class 11 chapter - CONIC SECTIONS​

Answer 1

Step-by-step explanation:

Given:

Centre of circle (2,-3)

eqn of first line -> 3x - 2y = 1 ___(1)

eqn of second line -> 4x + y = 27 ___(2)

multiply eqn (2) by 2 and adding with (1), we get

(8x + 2y) + (3x - 2y) = 54 + 1

11x = 55 => x = 5

putting value of x in eqn(2), we get

20 + y = 27

y = 7

point of interaction is (5,7)

eqn of circle is:

$( x - h) ^{2} + (y - k) ^{2} = r^{2} \\ (5 - 2) ^{2} + (7 + 3) ^{2} = r ^{2}$

$9 + 100 = r ^{2}$

$r^{2} = 109$

eqn of circle becomes:

$(x - 2)^{2} + (y + 3)^{2} = r^{2}$

$(x - 2)^{2} + (y + 3)^{2} = 109$