Question

slove please urgent!​

Answer 1

Answer:

$\huge \green{ \boxed{\sf \: a = 2}} \\ \\ \orange{ \boxed{\sf \:\: radius \: \: r = \sqrt{3} \: \: unit}} \:$

Step-by-step explanation:

We know that,

General equation of a circle is :

$\sf \: a {x}^{2} + b{y}^{2} + 2gx + 2fy + c = 0$

It will be a equation of circle if a = b

So we can say that,

If x and y are squared and the coefficient of x^2 and y^2 are same, then it is an equation of circle.

and its radius is :

$\sf \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ \sf \: \: and \: its \: center \: is \: ( - g, - f)$

Now given that,

$\sf \: 2 {x}^{2} + a {y}^{2} - 3x + 2y - 1 = 0$

If it is an equation of circle then the coefficient of x2 and y2 are same.

So,

$\huge \green{ \boxed{\sf \: a = 2}}$

So, equation is :

$\sf \: \: \: 2 {x}^{2} + 2 {y}^{2} - 3x + 2y - 1 = 0 \\ \sf \implies {x}^{2} + {y}^{2} - \frac{3}{2} x + y - \frac{1}{2} = 0 \: \: \: \red{ \{ \: divided \: by \: 2 \}}$

To find g and f, we will compair the given equation with general equation of circle

$\sf \: 2gx = - \frac{3}{2} x \: \: \: \therefore \: g = - \frac{3}{4} \: \: \: \: \pink{and} \\ \\ \sf \: 2fy = y \: \: \: \: \therefore \: f = \frac{1}{2} \: \: \: \: \: \: \pink{and \: also}\\ \\ \sf \: c = - \frac{1}{2} \\ \\ \bf \: now \: radius \: \\ \sf \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ \sf \: \: \: \: \: = \sqrt{ {( - \frac{3}{4} })^{2} + {( \frac{1}{2} })^{2} - ( - \frac{1}{2} ) } \\ = \sqrt{ \frac{9}{4} + \frac{1}{4} + \frac{1}{2} } \\ = \sqrt{ \frac{9 + 1 + 2}{4} } \\ = \sqrt{ \frac{12}{4} } \\ = \sqrt{3} \: unit \: \\ \\ \orange{ \boxed{\sf \: so \: radius \: \: r = \sqrt{3} \: \: unit}}$