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# Question

# Answer

(a) Since 3y =2x + 5 is normal to the circle, it must pass through the centre of the circle.

If equation of the circle is **(x – h) ^{2}+ (y – k)^{2} = r^{2}** , then 3k = 2h + 5

Given x = -2 and y = -1 are tangents to a circle, the centre of the circle must be equidistance from the line x = -2 and y = -1. The straight line that is equidistance ( m =1) from these two line consists of a point ( -2, -1) with equation of the line

y = x + c where c is a constant

y = x + c

at (-2, -1)

c =1

y = x + 1

k = h + 1

3( h + 1) = 2h + 5

3h + 3 = 2h + 5

h = 2

k = 3

So centre of the circle is (2, 3)

(b) If (2, 3 ) is the centre of the circle,

radius of the circle is 3 +1 = 4. ( Imagine distance from centre of circle vertically along y axis to y = -1)

So equation of the circle is **(x – 2) ^{2}+ (y – 3)^{2} = 4^{2}**