Question

How do you write the equation of the circle passing through (−2,1) and tangent to the line 3x − 2y = 6 at the point (4, 3)?

Answer 1

CnddkkgvekdkehgjofdlgchklrjjjjjjjrjrlrhvGMkk

Answer 2

Given that the circle passes through (-2,1).

A line 3x-2y =6 touches the circle at (4,3)

To write the equation of the circle:

We know that a circle is locus of all points equidistant from centre.

Let centre be (h,k)

Then (h,k) is equidistant from (-2,1) and (4,3)

In other words distance square would be equal.

$(h+2)^2+(k-1)^2 = (h-4)^2+(k-3)^2</p> <p>4h-2k+5 = -8h-6k+25</p><p>12h+4k =20 Or 3h+k = 5$

The slope of line joining (4,3) and (h,k) is $\frac{3-k}{4-h}$

since this is normal to tangent, slope = -1/slope of given line

= -2/3

2h+3k=17:

solving h =

Solving the two equations we get

h = 1.778 and k = -0.333

Radius = distance between (-2,1) and centre

= 4.01

Hence circle equation is

$(x-1.778)^2+(y+0.333)^2 = 4.01^2$