Question

Newbie here, nakakabaliw na po haha, paturo naman po! Maraming maraming salamat po talaga!

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

I have solved only two sub questions from each main question

rest do by yourself as self practice

Step-by-step explanation:

$II \: \: Q.1 \\ solution : \\ given \: equation \: of \: circle \: is \\ (x - \sqrt{2} ) {}^{2} + (y - \sqrt{7} ) {}^{2} = 10 \\ \\ comparing \: it \: with \: centre \: radius \\ form \: of \: circle \: \\ we \: get \\ h =g = \sqrt{2} \\ k = f = \sqrt{7} \\ r = \sqrt{10} \\ \\ we \: know \: that \\ r = \sqrt{g {}^{2} + f {}^{2} - c} \\ \\ \sqrt{10} = \sqrt{( \sqrt{2}) {}^{2} + ( \sqrt{7} ) {}^{2} - c } \\ \\ squaring \: both \: sides \\ we \: get \\ 10 = ( \sqrt{2} ) {}^{2} + ( \sqrt{7} ) {}^{2} - c \\ \\ = 2 + 7 - c \\ \\ c = 9 - 10 \\ \\ c = - 1 \\ \\ so \: we \: have \\ general \: form \: of \: circle \: as \\ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0 \\ \\ x {}^{2} + y {}^{2} + 2 \sqrt{2} .x + 2 \sqrt{7} .y + ( - 91) = 0 \\ \\ x {}^{2} + y {}^{2} + 2 \sqrt{2} x + 2 \sqrt{7} y - 91 = 0 \\ which \: is \: required \: \: generalised \: form \\ equation \: of \: circle$

$Q.2 \\ solution : \\ given \: equation \: of \: circle \: is \\ \\ (x - \frac{1}{2} \: ) {}^{2} + (y + \frac{1}{2} ) {}^{2} = \frac{169}{4} \\ \\ comparing \: with \\ centre \: radius \: form \: of \: circle \\ we \: get \\ \\ h = g = \frac{1}{2} \\ \\k = f= \frac{ - 1}{2} \\ \\ r = \frac{13}{2} \\ \\ we \: have \\ r = \sqrt{g {}^{2} + f {}^{2} - c } \\ \\ \frac{13}{2} = \sqrt{( \frac{1}{2} ) {}^{2} + ( \frac{ - 1}{2}) {}^{2} - c } \\ \\ squaring \: both \: sides \\ we \: get \\ \\ \frac{169}{4} = \frac{1}{4} + \frac{1}{4} - c \\ \\ \frac{169}{4} = \frac{1 + 1 - 4c}{4} \\ \\ 169 = 2 - 4c \\ \\ \\ 4c = - 167 \\ \\ c = \frac{ - 167}{4} \\ \\ now \: we \: know \: that \\ general \: form \: of \: circle \: is \: \\ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0 \\ \\ x {}^{2} + y {}^{2} + 2 \times \frac{1}{2} x + 2 \times ( \frac{ - 1}{2} y) - \frac{167}{4} = 0 \\ \\ x {}^{2} + y {}^{2} + x - y - \frac{167}{4} = 0 \\ \\ which \: is \: required \: \: general \: equation \\ of \: circle$

$III \: \: Q.1 \\ solution : \\ given \: \: equation \: of \: circle \: is \\ x {}^{2} + y {}^{2} - 12x + 4y + 15 = 0 \\ \\ comparing \: it \: with \\ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0 \\ we \: get \\ \\ 2g = - 12 \\ ∴ \: g = h = - 6 \\ \\ 2f = 4 \\ ∴ \: f = k = 2 \\ \\ c = 15 \\ \\∴ \: coordinates \: of \: center = ( - g, - h) \\ = (6, - 2) \\ \\ we \: know \: that \\ r = \sqrt{g {}^{2} + f {}^{2} - c } \\ \\ = \sqrt{( - 6) {}^{2} + (2) {}^{2} - 15} \\ \\ = \sqrt{36 + 4 - 15} \\ \\ = \sqrt{36 - 11} \\ \\ = \sqrt{25} \\ \\ r = 5 \\ \\ so \: now \: \: we \: have \\ (x - h) {}^{2} + (y - k) {}^{2} = r {}^{2} \\ \\ (x - 6 ) {}^{2} + (y - ( - 2)) {}^{2} = (5) {}^{2} \\ \\ ∴ \: (x - 6) {}^{2} + (y + 2) {}^{2} = 25$