Find the equation of a circle whose centre is (3, –1)
Question:

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

Solution:

Given equation of the chord is,

$2 x-5 y+18=0$

$5 y=2 x+18$

$y=\frac{2}{5} x+\frac{18}{5}$

As we have, $y=m x+C$

Where, $\mathrm{m}$ is the slope of the line,

$m=\frac{2}{5}$

Slope of the line perpendicular to the chord,

$m^{\prime}=-\frac{5}{2}$

As the product of slope of perpendicular lines $=-1$,

$y-y_{1}=m^{\prime}\left(x-x_{1}\right)$

$y-(-1)=-\frac{5}{2}(x-3)$

$2 y+2=-5 x+15$

$5 x+2 y=13$

$.2$ [Equation of line passing from centre and cutting the chord]

Solving both the equations,

$2 x-5 y=-18 \& 5 x+2 y=13$

Multiplying the equation $1 \&$ equation 2 by $2 \& 5$ respectively, we get

$4 x-10 y=-36$

$25 x+10 y=65$

$29 x=29$

$x=1$

$2(1)-5 y=-18$

$2-5 y+18=0$

$5 y=20$

$y=4$

Point of intersection at chord and radius $=(1,4)$ Distance between the point of intersection \& centre

Distance formula $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(1-3)^{2}+(4-(-1))^{2}}$

$=\sqrt{(-2)^{2}+(4+1)^{2}}$

$=\sqrt{4+(5)^{2}}$

$=\sqrt{4+25}$

$=\sqrt{2} 9$ units

Using Pythagoras Theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

= (3)2 + (√29)2 = 29 + 9

= √38

Hypotenuse = √38 units (radius)

Since, the radius bisects the chord into two equal halves,

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)2 + (y – k)2 = r2

(x – 3)2 + (y – (-1))2 = (√38)2

x2 – 6x + 9 + (y + 1)2 = 38

x2 – 6x + y2 + 2y + 1 + 9- 38 = 0

x2 – 6x + y2 + 2y – 28 = 0

Hence, the required equation of the circle is x2 – 6x + y2 + 2y – 28 = 0.

 

 

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