Answer:
A tangent to a circle is a straight line which touches the circle at only one point. Therefore, if [tex]x=2y+5[/tex] is a tangent of [tex]x^2+y^2=5[/tex] then there will be one point of intersection.
To find the point of intersection, substitute [tex]x=2y+5[/tex] into [tex]x^2+y^2=5[/tex] and solve for y:
[tex]\implies (2y+5)^2+y^2=5[/tex]
[tex]\implies 4y^2+20y+25+y^2=5[/tex]
[tex]\implies 5y^2+20y+20=0[/tex]
[tex]\implies y^2+4y+4=0[/tex]
[tex]\implies (y+2)^2=0[/tex]
[tex]\implies y+2=0[/tex]
[tex]\implies y=-2[/tex]
Substitute found value of y into [tex]x=2y+5[/tex] and solve for x:
[tex]\implies x=2(-2)+5=1[/tex]
Therefore, there is one point of intersection at (1, -2), thus proving that the straight line equation is a tangent to the circle.
