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Answer:

A chord of a circle, x - y = 0, creates the angle of 45° and 135° with the circumference of the circle x² + y² - 2x + 6y - 6 = 0.

Step-by-step explanation:

Given equation of circle is x² + y² - 2x + 6y - 6 = 0 ---(1)

So,

[tex]\sf\: Center\:of\:circle = \bigg( - \dfrac{1}{2} \: coeff. \: of \: x, \: - \dfrac{1}{2} \: coeff. \: of \: y \bigg) \\ [/tex]

[tex]\sf\: Center\:of\:circle = \bigg( - \dfrac{1}{2} \: ( - 2), \: - \dfrac{1}{2} \: (6) \bigg) \\ [/tex]

[tex]\implies\sf\:Center\:of\:circle = (1, - 3) \\ [/tex]

Now,

[tex]\sf\: Radius\:of\:circle = \sqrt{ \bigg( - \dfrac{1}{2} \: coeff. \: of \: x\bigg)^{2} + {\bigg(- \dfrac{1}{2} \: coeff. \: of \: y \bigg) }^{2} - constant} \\ [/tex]

[tex]\sf\: Radius\:of\:circle = \sqrt{ {(1)}^{2} + {( - 3)}^{2} - ( - 6)} \\ [/tex]

[tex]\sf\: Radius\:of\:circle = \sqrt{ 1 + 9 + 6} \\ [/tex]

[tex]\sf\: Radius\:of\:circle = \sqrt{ 16} \\ [/tex]

[tex]\implies\sf\:Radius\:of\:circle = 4 \: units \\ [/tex]

Now, Let's evaluate the point of intersection of chord of a circle, x - y = 0 with the circle x² + y² - 2x + 6y - 6 = 0.

On substituting the value of x = y in equation (1), we get

[tex]\sf\: {x}^{2} + {x}^{2} - 2x + 6x - 6 = 0 \\ [/tex]

[tex]\sf\: 2{x}^{2} + 4x - 6 = 0 \\ [/tex]

[tex]\sf\: 2({x}^{2} + 2x - 3) = 0 \\ [/tex]

[tex]\sf\: {x}^{2} + 2x - 3 = 0 \\ [/tex]

[tex]\sf\: {x}^{2} + 3x - x - 3 = 0 \\ [/tex]

[tex]\sf\: x(x + 3) - 1(x + 3) = 0 \\ [/tex]

[tex]\sf\: (x + 3) (x - 1) = 0 \\ [/tex]

[tex]\implies\sf\:x = - 3 \: \: or \: \: x = 1 \\ [/tex]

Thus,

[tex]\implies\sf\:y= - 3 \: \: or \: \: y = 1 \\ [/tex]

So, the point of intersection of chord of a circle, x - y = 0 with the circle x² + y² - 2x + 6y - 6 = 0 are (- 3, - 3) and (1, 1)

Let assume that the point of intersection be represented as A and B respectively such that A be (- 3, -3) and B be (1, 1).

Now, Consider

[tex]\sf\: {AO}^{2} = {4}^{2} = 16 \\ [/tex]

[tex]\sf\: {OB}^{2} = {4}^{2} = 16 \\ [/tex]

[tex]\sf\: {AB}^{2} = {( - 3 - 1)}^{2} + {( - 3 - 1)}^{2} = 16 + 16 = 32\\ [/tex]

So, we have

[tex]\implies\sf\: {AO}^{2} + {OB}^{2} = {AB}^{2} \\ [/tex]

[tex]\implies\sf\: \angle \: AOB = {90}^{ \circ} \\ [/tex]

Let assume C and D be two point on the circumference of a circle on major arc and minor arc.

We know, Angle subtended at the centre by an arc is double the angle subtended on the circumference by the same arc.

[tex]\implies\sf\: \angle \: ACB = \dfrac{1}{2} \angle \:AOB = \dfrac{1}{2} \times {90}^{ \circ} = 45^{ \circ} \\ [/tex]

Now, ADBC is a cyclic quadrilateral. We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.

[tex]\sf\: \angle \:ADB + \angle \:ACB = 180^{ \circ} \\ [/tex]

[tex]\sf\: \angle \:ADB + 45^{ \circ} = 180^{ \circ} \\ [/tex]

[tex]\sf\: \angle \:ADB = 180^{ \circ} - 45^{ \circ}\\ [/tex]

[tex]\implies\bf\: \angle \:ADB = 135^{ \circ}\\ [/tex]

Hence, A chord of a circle, x - y = 0, creates the angle of 45° and 135° with the circumference of the circle x² + y² - 2x + 6y - 6 = 0.