Question

Verodharti shabdh marathi hadu Hadu 6th standard

Answer 1

Answer:

Radius, r = 4, and center (h, k) = (-2, 3).

We know that the equation of a circle with centre (h, k) and radius r is given as

→ (x – h)² + (y – k)² = r² \: \: \: ….(1)→(x–h)²+(y–k)²=r²….(1)

Now, substitute the radius and center values in (1), we get

Therefore, the equation of the circle is

→ (x + 2)²+ (y – 3)² = (4)²→(x+2)²+(y–3)²=(4)²

→ x²+ 4x + 4 + y² – 6y + 9 = 16→x²+4x+4+y²–6y+9=16

Now, simplify the above equation, we get:

→ x² + y²+ 4x – 6y – 3 = 0→x²+y²+4x–6y–3=0

Thus, the equation of a circle with center (-2, 3) and radius 4 is :

→ x² + y²+ 4x – 6y – 3 = 0→x²+y²+4x–6y–3=0