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According to the most recent employment data shared by MIT School of Engineering, out of the entire class of 2022, this school continues to demonstrate exceptional career outcomes, as the MIT School of Engineering placement rate for master's grads in that year was 88.5% and 88% for bachelor's graduates. The starting salary of bachelor's grads in that same year was USD 126,841 (INR 1.08 Cr., ) and it was USD 166,300 (INR 1.42 Cr.) for the postgraduate students.

Among the top companies that hire the MIT Engineering grads are Google, Apple, Amazon, OpenAI, Bain & Company, JP Morgan, Goldman Sachs, Boeing, Caterpillar, NVidia, etc.

11. Let the equation of the circle be.

(x – h)² + (y – k)² = r² -----(i)

Since the circle passes through (2, 3) and (–1, 1),

Putting x = 2 and y = 3 in (i),

(2 – h)² + (3 – k)² = r²---------(ii)

Putting x = –1 and y = 1 in (i),

(–1 – h)² + (1 – k)² = r²

Equating equation (ii) and (iii), We get:–

(2 – h)² + (3 – k)² = (–1 – h)² + (1 – k)²

2² + h² – 2.2.h + 3² + k² – 2.3.k = (–1)² + h² – 2.(–1).h + 1² + k² – 2(1)(k)

4 + h² – 4h + 9 + k² – 6k = 1 + h² + 2h + 1 + k² – 2k

13 – 4h – 6k = 2 + 2h – 2k

–4h – 2h – 6k + 2k = 2 – 13

–6h – 4k = 11

–(6h + 4k) = 11

6h + 4k = 11-------(iv)

Since the centre of the circle is on the line x – 3y – 11 = 0

? h – 3k = 11

(h – 3k = 11) × 6 = 6h – 18k = 66-----(v)

Subtract (v) – (iv),

$\begin{array}{l}\cancel{6h}-18k=66\\ \underset{\_}{\left(\cancel{6h}\pm 4k=-11\right)}\end{array}$

– 22k = 55

k = $\frac{-55}{22}=\frac{-5}{22}$

Putting the value of k in (iv), we get

6h + $\cancel{\stackrel{\mathrm{2/4}}{}}$ × $\left(-\frac{55}{\underset{11}{\cancel{22}}}\right)=11$

6h + $\left(-\frac{110}{11}\right)=11$

6h = 11 + $\frac{\cancel{\stackrel{10}{110}}}{\underset{1}{\cancel{11}}}$

6h = 11 + 10

h = $\frac{{\cancel{21}}^7}{\cancel{\underset{2}{6}}}=\frac{7}{2}$

Putting values of h and k in (ii), we get

${\left(2-\frac{7}{2}\right)}^2+{\left(3+\frac{5}{2}\right)}^2=r^2$

${\left(\frac{4-7}{2}\right)}^2+{\left(\frac{6+5}{2}\right)}^2=r^2$

${\left(\frac{-3}{2}\right)}^2+{\left(\frac{11}{2}\right)}^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$\frac{9+121}{4}=r^2$

r² = $\frac{\stackrel{65}{\cancel{130}}}{\cancel{\underset{2}{4}}}$

r² = $\frac{65}{2}$

Than the equation of the circle is,

${\left(x-\frac{7}{2}\right)}^2+{\left(y+\frac{5}{2}\right)}^2=\frac{65}{2}.$

$\Rightarrow x{}_2+y^2-7x+5y-14=0$

MIT School of Engineering remains a top choice for Engineering students all over the world. With it's motto of 'mind and hand',  this school is lleading all the rankings and charts. 

According to the most recent employment data shared by MIT School of Engineering, out of the entire class of 2022, this school continues to demonstrate exceptional career outcomes, as the MIT School of Engineering placement rate for master's grads in that year was 88.5% and 88% for bachelor's graduates. The starting salary of bachelor's grads in that same year was USD 126,841 (INR 1.08 Cr., ) and it was USD 166,300 (INR 1.42 Cr.) for the postgraduate students.

Among the top companies that hire the MIT Engineering grads are Google, Apple, Amazon, OpenAI, Bain & Company, JP Morgan, Goldman Sachs, Boeing, Caterpillar, NVidia, etc.

10. Let the equation of the circle be,

(x – h)² + (y – k)² = r² - (i)

Since the circle passes through (4, 1) and (6, 5)

Putting x = 4 and y = 1 in (i),

(4 – h)² + (1 – k)² = 2² - (ii)

Putting x = 6 and y = 5 in (i),

(6 – h)² + (5 – k)² = r²- (iii)

Equating equation (ii) and (iii), We get.

(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

4² + h² – 2.4.h + 1² + k² – 2.1.k = 6² + h² – 2.6.h + 5² + k² – 2.5.k

16 + h² – 8h + 1 + k² – 2k = 36 + h² – 12h + 25 + k² – 10k

17 + h² – 8h + k² – 2k = 61 + h²  – 12h + k² – 10k

–8h + 12h – 2k + 10k = 61 – 17

4h + 8k = 44

4 (h + 2k) = 44

h + 2k = $\frac{44}{4}$ = 11- (iv)

Since the centre of the circle passes through 4x + y = 16

? 4h + k = 16

(4h + k = 16) × 2

8h + 2k = 32- (v)

Subtract (v) – (iv).

8h + 2k = 32

– (h + 2k = 11)

7h = 21

h = $\frac{21}{7}$ = 3

Putting h = 3 in (iv), we get

3 + 2k = 11

2k = 11 – 3

k = $\frac{8}{2}$ = 4

Putting the values of h and k in, (ii)

(4 – 3)² + (1 – 4)² = r²

1² + (–3)² = r²

1 + 9 = r²

r = √10$$

Thus the equation of the circle is,

(x – 3)² + (y – 4)² = ${(\surd 10)}^2$

(x – 3)² + (y – 4)² = 10

x²$+$$y^2-6x-8y+15=0$

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