Question

29. For what value of k will (k - 5)x2+ 2ky2-5x +
6y - 3 = 0 represents a circle?
(1 Point)
- 5
Ол
10​

Answer 1

The value of k = - 5 so that (k - 5)x² + 2ky² - 5x + 6y - 3 = 0 represents a circle.

Given :

The equation (k - 5)x² + 2ky² - 5x + 6y - 3 = 0

To find :

The equation (k - 5)x² + 2ky² - 5x + 6y - 3 = 0 represents a circle when k is

• - 5

• 10

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

(k - 5)x² + 2ky² - 5x + 6y - 3 = 0

Step 2 of 2 :

Find the value of k

We know that the equation ax² + by² + 2gx + 2fy + c = 0 represents a circle if

Coefficient of x² = Coefficient of y²

i.e a = b

Now the equation (k - 5)x² + 2ky² - 5x + 6y - 3 = 0 represents a circle

So we have

Coefficient of x² = Coefficient of y²

⇒ k - 5 = 2k

⇒ 2k = k - 5

⇒ 2k - k = - 5

⇒ k = - 5

So the value of k = - 5

Hence the correct option is - 5

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Answer 2

Answer:

k = -5

Step-by-step explanation:

Given:- $(k-5)x^{2}+2ky^{2}-5x+6y-3=0$

To Find:- Value of k

Solution:-

As we know, The equation of a circle is  $ax^{2} +by^{2}+2gx+2fy+c=0$ if and only if coefficient of $x^{2}$ = coefficient of $y^{2}$

∴ (k-5) = 2k

⇒ 2k-k = -5

⇒ k = -5.

Hence, correct option is -5.

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