Question

my teacher directly marked the point (-4,3),(4,3),(5,0),(0,-5)
on the circle how did he come to know about this how did he analyzed no equation no centre no radius was given​

Answer 1

$\huge\color{red}{\underline{\underline{question\::}}}$

my teacher directly marked the point (-4,3),(4,3),(5,0),(0,-5)on the circle how did he come to know about this how did he analyzed no equation no centre no radius was given.

$\huge\color{blue}{\underline{\underline{answer\::}}}$

x² + y² - 2x - 4y = 0

$\huge\color{pink}{\underline{\underline{explanation\::}}}$

Let (x−h)² + (y−k)² = r² …………..(1)

where (h, k) is the centre of the circle and r is the radius.

Let us draw a line A(2,0) and B(4,0). If the centre lies outside the line AB then it is a chord.

The diameter of the circle is bigger than the chord. Suppose AB is the diameter of the circle then the centre of the circle must be a midpoint of AB.

∴ Using section formula

• h = (2+0)/2 = 1

• k = (4+0)/2 = 2

∴ Coordinates of the centre=(h,k) =(1,2)

Equation (1) becomes,

(x−1)² + (y - 2)² = r² …………….(2)

Since (2) passes through (2,0), equation (2) can be written as,

(2−1)² +(0−2)² = r²

• 1² + 2² = r²

• 1+4 = r²

• r² = 5

• r = √5

Equation of the circle with minimum radius is

• ∴(x−1)² + (y−2)² = 5

• ⇒x² +1−2x+y² + 4 − 4y = 5

• ⇒x² + y² - 2x - 4y + 5 - 5 = 0

• ⇒x² + y² - 2x - 4y = 0

$\huge\color{yellow}{\underline{\underline{so \: the \: final \: equation \: is \: \::}}}$

x² + y² - 2x - 4y = 0

Answer 2

Step-by-step explanation:

$x² + y² - 2x - 4y = 0</p><p></p><p>\huge\color{pink}{\underline{\underline{explanation\::}}}explanation:</p><p></p><p>Let (x−h)² + (y−k)² = r² …………..(1)</p><p></p><p>where (h, k) is the centre of the circle and r is the radius.</p><p></p><p>Let us draw a line A(2,0) and B(4,0). If the centre lies outside the line AB then it is a chord.</p><p></p><p>The diameter of the circle is bigger than the chord. Suppose AB is the diameter of the circle then the centre of the circle must be a midpoint of AB.</p><p></p><p>∴ Using section formula</p><p></p><p>h = (2+0)/2 = 1</p><p></p><p>k = (4+0)/2 = 2</p><p></p><p>∴ Coordinates of the centre=(h,k) =(1,2)</p><p></p><p>Equation (1) becomes,</p><p></p><p>(x−1)² + (y - 2)² = r² …………….(2)</p><p></p><p>Since (2) passes through (2,0), equation (2) can be written as,</p><p></p><p>(2−1)² +(0−2)² = r²</p><p></p><p>1² + 2² = r²</p><p></p><p>1+4 = r²</p><p></p><p>r² = 5</p><p></p><p>r = √5</p><p></p><p>Equation of the circle with minimum radius is</p><p></p><p>∴(x−1)² + (y−2)² = 5</p><p></p><p>⇒x² +1−2x+y² + 4 − 4y = 5</p><p></p><p>⇒x² + y² - 2x - 4y + 5 - 5 = 0</p><p></p><p>⇒x² + y² - 2x - 4y = 0</p><p></p><p>\huge\color{yellow}{\underline{\underline{so \: the \: final \: equation \: is \: \::}}}sothefinalequationis:</p><p></p><p>x² + y² - 2x - 4y = 0</p><p></p><p>$