Question

Q7.Findtheequationofthecirclewhichpassesthroughthepoints(1,-2)and(4,-3)andhasits

centreon

theline3x+4y=7.

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Answer 1

The general equation of a circle whose centre is (h,k) and radius r is

(x - h)² + (y - k)² = r²

Since the centre (h,k) of the circle is on the line 3x + 4y = 7, it must satisfy equation of line

3h + 4k = 7....(i)

We can also say that distance between (h,k) and (1,-2) is equal to distance between (h,k) and (4,-3) because the radii is same.

Hence by distance formula

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

${(h + 1)^2 + (k-2)^2} = {(h + 4)^2 + (k-3)^2}$

${(h + 1)^2 - (h+4)^2} = (k-3)^2 - (k-2)^2$

$(h+1+h+4)(h+1-h-4) = (k-3+k-2)(k-3-k+2)$

$(2h+5)(-3) = (2k-5)(-1)$

$6h+15 = 2k-5$

$6h-2k = -20$

12h - 4k = -40....(ii)

Solving equation (i) and (ii)

h = -11/5

k = 17/5

radii (from distance formula) = √3.4

So the equation of circle is

(x + 11/5)² + (k - 17/5)² = 3.4