Question

cance, a LCR series circui, the current and ent are not in plase with salah 14. Aseta (Al Uelike electric forces and gravitational forte calea face baslistage Reason (R! Nocles forces de not obcy inese square la SECTION-B Questions 15 and 16 are Case Study based questions and be compbry. Aseçim 4 suku-s bun acil question. Each question carries i es 15. Usung Galzones as an Ammer and a Voltran & galvromer is a device wet o dezer can creat, the cannot as such be used as an aeter to more come in a given circuit . This is because alunomener vey sensitive devices, it gives a full-scale dellection for a current of the order of A. Moreover, or muruniig suren, die galvanomca has to be connected in scris, and it has a large rece, discharge de valores de cine the gammeter To overcome these difficulties, we corner a small tituee Ry, culadatura ren sanca, in para coil, so that most of the current passes through the show. Now we gL VIOLET & ver, it has to be someone is parallel with the circuit element zross stick we need ours Morose, i mudru vary amall TCE otherwise will repreciably change thevoluge which we we meaning to come is a large resudarce Recomendan series with the galvanometer​

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