for the function f(x) = 3(x - 1)^2 + 2, identify the vertex, domain and range
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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
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