Question

the eq. of circle whose diameter lies on 2x+3y=3 and 16x-y=4 and which passes through (4,6)is
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Answer 1

Answer:

5x2+5y2−3x−8y = 200

Step-by-step explanation:

The equation of diameters of the circle are;

2x+3y = 3 ...(i)16x−y = 4 ...(ii)2x+3y = 3 ...(i)16x-y = 4 ...(ii)

Multiplying (i) by 8 and then subtracting (ii) from (i) we get;

16x+24y−16x+y = 24−4⇒25y = 20⇒y = 2025 = 4516x+24y-16x+y = 24-4⇒25y = 20⇒y = 2025 = 45

And from (ii) we have;

16x−45 = 4⇒16x = 4+45 ⇒16x = 245⇒x = 245×16 = 31016x-45 = 4⇒16x = 4+45 ⇒16x = 245⇒x = 245×16 = 310

So the centre of the circle is (310, 45)310, 45

We know that the equation of the circle with the centre (h,k)h,k and radius r is (x−h)2+(y−k)2 = r2x-h2+y-k2 = r2

So the equation of the required circle with centre (310, 45)310, 45 and radius r is (x−310)2+(y−45)2 = r2x-3102+y-452 = r2

Since this circle passes through the point (4, 6), so we have;

(4−310)2+(6−45)2 = r2⇒(3710)2+(265)2 = r2⇒r2 = 1369100+67625⇒r2 = 4073100  ...(iii)4-3102+6-452 = r2⇒37102+2652 = r2⇒r2 = 1369100+67625⇒r2 = 4073100  ...(iii)

So the equation of the circle is given by;

(x−310)2+(y−45)2 = r2⇒(x−310)2+(y−45)2 = 4073100     {using (iii)}⇒x2+9100−6x10+y2+1625−8y5 = 4073100⇒100x2+9−60x+100y2+64−160y100 = 4073100⇒100x2+100y2−60x−160y+73 = 4073⇒100x2+100y2−60x−160y = 4000⇒5x2+5y2−3x−8y = 200x-3102+y-452 = r2⇒x-3102+y-452 = 4073100     {using (iii)}⇒x2+9100-6x10+y2+1625-8y5 = 4073100⇒100x2+9-60x+100y2+64-160y100 = 4073100⇒100x2+100y2-60x-160y+73 = 4073⇒100x2+100y2-60x-160y = 4000⇒5x2+5y2-3x-8y = 200

Therefore the equation of the circle is 5x2+5y2−3x−8y = 200

Answer 2

Let the centre of the circle be at (h, k)

Then (h, k) lies on both the straight lines 2x + 3y = 3 and 16x - y = 4.

Thus we find:

2h + 3k = 3 ....(i)

16h - k = 4 .....(ii)

Multiplying (ii) by 3 and adding with (i), we get:

48h - 3k + 2h + 3k = 12 + 3

or, 50h = 15

or, h = 3/10

Putting h = 3/10 in (ii), we get:

16 (3/10) - k = 4

or, 48/10 - k = 4

or, k = 48/10 - 4

or, k = 8/10

So, (3/10, 8/10) is the centre of the circle.

The circle also passes through the point (4, 6). Thus its radius is

= √{ (4 - 3/10)² + (6 - 8/10)² } units

= √{ (37/10)² + (52/10)² } units

= √(1369/100 + 2704/100) units

= (√4073)/10 units

Therefore the equation of the circle is

(x - 3/10)² + (y - 8/10)² = {(√4073)/10}²

or, x² - 6x/10 + 9/100 + y² - 16y/10 + 64/100

= 4073/100

or, x² + y² - 6x/10 - 16y/10 = 40

or, 10x² + 10y² - 6x - 16y = 400

or, 5x² + 5y² - 3x - 8y = 200.

Answer: The required circle is

5x² + 5y² - 3x - 8y = 200.