Question

13. Radius of circle (x-5)(x-1) + (y-7)(y-
4) = O is​

Answer 1

Answer:

[ x-x1 ] [ x- x2] + [ y - y1] [ y-y2] = 0

[x1 y1 ] and [x2 y2]

[ 5,7 ] and [1,4]

r = 1/2

root [ 5-1]^2 +[ 7 - 4]^2

=1/2  root 4^2 + 3^2 = 5/2

hope it  helps

Answer 2

Answer:

Radius of circle (x-5)(x-1) + (y-7)(y-4) = 0 is​ $\frac{5}{2}$

Step-by-step explanation:

Given:

Equation of the circle (x-5)(x-1) + (y-7)(y-4) = 0

To Find:

The radius of the circle

Chapter:

Circle

Concept:

The standard form of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$

Center = (-g,f)

Radius =$\sqrt{g^{2} +f^{2} -c}$

Solution:

(x - 5) (x – 1) + (y – 7) (y – 4) = 0 (Given)

$= x^2 - 6x + 5 + y^2 - 11y + 28 = 0$

$=x^2 + y^2 - 6x - 11y + 33 = 0$

Compared with a standard form of a circle

$x^2 + y^2 + 2gx + 2fy + c = 0$

Therefore,

• 2gx = -6x
• 2fy = -11y

Thus, g = -3 and f = $-\frac{11}{2}$

On this basis,

c = 33

Further, Center = (-g,f)

Now as we have all the values we need, we will substitute them in the equation of the radius to get the final answer:

Radius =$\sqrt{g^{2} +f^{2} -c}$

= $\sqrt{(-3)^{2} +(-\frac{11}{2}) ^{2} -33}$

$= \sqrt{9 +(\frac{121}{4} )-33}$

$=\sqrt{\frac{25}{4} }$

Radius = $\frac{5}{2}$

Conclusion:

Radius of circle (x-5)(x-1) + (y-7)(y-4) = 0 is $\frac{5}{2}$

#SPJ2