Answer:

[tex] \huge \bold \red{hello}[/tex]

Solution :

Given the equation to the circle is ~

(x-a)(x-b)+(y-b)(y-d) = 0

or, x²-x(a+b)+ab+y²-x(c+c)+cd=0

or, x²-2.x. a+b/2 + (a+b)²/4-ab + (c+d)/4 - CD

OR,

(x-a+b/2)²+(y-c+d/2)²=(a-b)²/4 + (c-d)²/4.

[tex]from \: this \: equations \: it \: is \: clear \: that \: the \: circle \: is \: [/tex]

(a+b/2),(c+d/2)

Thanks have a Nice day :) ❤️

Verified answer

Answer:

Hello Mathematician !!!

There are two methods to solve this equation.

I gonna tell you the easiest ones ｡◕‿◕｡

Solution:-

Using Pythagoras theorem

[tex] \tiny[{(x - a)}^{2} + {(y - b)}^{2}] + [ {(x - c)}^{2} + {(y - d)}^{2} ] = {(a - c) ^{2} + {(b - d)}^{2} }\\ [/tex]

or (noting that the squares of constants cancel and dividing though by 2)

[tex] \tiny {x}^{2} - ax - cx + {y}^{2} - bx - dx = - ac - bd [/tex]

from which the conclusion follows.

[But the other way using the scalar/dot product is neater and shows more insight]

Small experience Story:-

I experienced this type of maths in my tests where my teachers used to cut my marks for one reason. when we go to them to ask why my marks have been removed?? they asked us a question. " How can we know that the square of the constant terms are cancelled??

Answer:-

When you expand [tex] {a}^{2} [/tex] on the left when you expand is clearly going to cancel with the [tex] {a}^{2} [/tex] on the right and likewise for

[tex] {b}^{2} , {c}^{2} , {d}^{2}: [/tex]if you see it right, it is obvious that they appear once on each side. On the other hand you get

[tex] {2x}^{2} [/tex] and [tex] {2y}^{2} [/tex] because the terms with those squares are on the same side - and the cross terms in the expansions all have a factor 2 as well.

Hope it helps you from my side!!

Keep Learning and solving!!(•‿•)

Description:-

• Well if you don't mind, can you help me with my math question (latest). Actually i need the dual of your and mathdude sir's answer. Your answer and mathdude sir's answer are different. I want to see two processes.

Thank you!!